{
  "index": "1939-B-3",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT",
    "COMB"
  ],
  "difficulty": "",
  "question": "10. Given the power-series\n\\[\na_{0}+a_{1} x+a_{2} x^{2}+\\cdots\n\\]\nin which\n\\[\na_{n}=\\left(n^{2}+1\\right) 3^{n}\n\\]\nshow that there is a relation of the form\n\\[\na_{n}+p a_{n+1}+q a_{n+2}+r a_{n+3}=0\n\\]\nin which \\( p, q, r \\) are constants independent of \\( n \\). Find these constants and the sum of the power-series.",
  "solution": "First Solution. The desired relation is\n\\[\n\\begin{array}{c}\n\\left(n^{2}+1\\right) 3^{n}+p\\left((n+1)^{2}+1\\right) 3^{n+1} \\\\\n+q\\left((n+2)^{2}+1\\right) 3^{n+2}+r\\left((n+3)^{2}+1\\right) 3^{n+3}=0\n\\end{array}\n\\]\nwhich is equivalent to\n\\[\n\\begin{array}{c}\nn^{2}(1+3 p+9 q+27 r)+n(6 p+36 q+162 r) \\\\\n\\quad+(1+6 p+45 q+270 r)=0\n\\end{array}\n\\]\n\nEquation (1) holds for all \\( n \\) if and only if\n\\[\n\\begin{array}{r}\n1+3 p+9 q+27 r=0 \\\\\np+6 q+27 r=0 \\\\\n1+6 p+45 q+270 r=0\n\\end{array}\n\\]\n\nThese linear equations have the solution \\( p=-1, q=\\frac{1}{3}, r=-\\frac{1}{27} \\), so\n\\[\na_{n}-a_{n+1}+\\frac{1}{3} a_{n+2}-\\frac{1}{27} a_{n+3}=0\n\\]\n\nLet \\( S(x)=a_{0}+a_{1} x+a_{2} x^{2}+\\cdots \\). Proceeding formally, we have\n\\[\n\\begin{array}{rlrl}\nx^{3} S(x) & = & a_{0} x^{3}+a_{1} x^{4}+\\cdots+a_{n-3} x^{n}+\\cdots \\\\\np x^{2} S(x) & = & p a_{0} x^{2}+p a_{1} x^{3}+p a_{2} x^{4}+\\cdots+p a_{n-2} x^{n}+\\cdots \\\\\nq x S(x) & = & q a_{0} x+q a_{1} x^{2}+q a_{2} x^{3}+q a_{3} x^{4}+\\cdots+q a_{n-1} x^{n}+\\cdots \\\\\nr S(x) & =r a_{0}+r a_{1} x+r a_{2} x^{2}+r a_{3} x^{3}+r a_{4} x^{4}+\\cdots+r a_{n} x^{n}+\\cdots\n\\end{array}\n\\]\n\nWhen we sum these we get\n\\[\nS(x)\\left[x^{3}+p x^{2}+q x+r\\right]=\\left(p a_{0}+q a_{1}+r a_{2}\\right) x^{2}+\\left(q a_{0}+r a_{1}\\right) x+r a_{0}\n\\]\n\nMultiplying through by -27 , we obtain\n\\[\nS(x)\\left[1-9 x+27 x^{2}-27 x^{3}\\right]=1-3 x+18 x^{2}\n\\]\nand therefore\n\\[\nS(x)=\\frac{1-3 x+18 x^{2}}{(1-3 x)^{3}}\n\\]\n\nUsing the ratio test we conclude that the series converges for \\( |x|<\\frac{1}{3} \\); hence the formal manipulations above are valid for these values of \\( x \\).\n\nSecond Solution. Let \\( b_{n}=a_{n} / 3^{n}=n^{2}+1 \\). Then\n\\[\n\\begin{aligned}\n\\Delta b_{n} & =b_{n+1}-b_{n}=2 n+1, \\quad \\Delta^{2} b_{n}=b_{n+2}-2 b_{n+1}+b_{n}=2 \\\\\n\\Delta^{3} b_{n} & =b_{n+3}-3 b_{n+2}+3 b_{n+1}-b_{n}=0\n\\end{aligned}\n\\]\n\nSo\n\\[\n\\frac{a_{n+3}}{3^{n+3}}-\\frac{3 a_{n+2}}{3^{n+2}}+\\frac{3 a_{n+1}}{3^{n+1}}-\\frac{a_{n}}{3^{n}}=0\n\\]\nwhence\n\\[\na_{n}-a_{n+1}+(1 / 3) a_{n+2}-(1 / 27) a_{n+3}=0\n\\]\n\nSince\n\\[\nn^{2}+1=(n+1)(n+2)-3(n+1)+2\n\\]\nwe have\n\\[\n\\begin{aligned}\n\\Sigma b_{n} y^{n} & =\\Sigma(n+1)(n+2) y^{n}-3 \\Sigma(n+1) y^{n}+2 \\Sigma y^{n} \\\\\n& =\\frac{2}{(1-y)^{3}}-\\frac{3}{(1-y)^{2}}+\\frac{2}{1-y} \\\\\n& =\\frac{1-y+2 y^{2}}{(1-y)^{3}}\n\\end{aligned}\n\\]\nprovided \\( |y|<1 \\). Replace \\( y \\) by \\( 3 x \\).\n\\[\n\\Sigma a_{n} x^{n}=\\frac{1-3 x+18 x^{2}}{(1-3 x)^{3}}\n\\]\nprovided \\( |x|<\\frac{1}{3} \\).\nRemark. We could assume that the problem is concerned with the ring of formal power series. In that case, \\( (1-3 x) \\) has an inverse in the ring and our result is that\n\\[\n\\sum a_{n} x^{n}=\\left(1-3 x+18 x^{2}\\right)(1-3 x)^{-3}\n\\]",
  "vars": [
    "x",
    "y",
    "n",
    "S",
    "a_0",
    "a_1",
    "a_2",
    "a_3",
    "a_4",
    "a_n",
    "a_n+1",
    "a_n+2",
    "a_n+3",
    "a_n-3",
    "a_n-2",
    "a_n-1",
    "b_n",
    "b_n+1",
    "b_n+2",
    "b_n+3"
  ],
  "params": [
    "p",
    "q",
    "r"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "y": "variabley",
        "n": "indexvar",
        "S": "seriesfunc",
        "a_0": "coefzero",
        "a_1": "coefone",
        "a_2": "coeftwo",
        "a_3": "coefthree",
        "a_4": "coeffour",
        "a_n": "coefgeneral",
        "a_n+1": "coefgennext",
        "a_n+2": "coefgenplus2",
        "a_n+3": "coefgenplus3",
        "a_n-3": "coefgenminus3",
        "a_n-2": "coefgenminus2",
        "a_n-1": "coefgenminus1",
        "b_n": "altcoefgeneral",
        "b_n+1": "altcoefgennext",
        "b_n+2": "altcoefgenplus2",
        "b_n+3": "altcoefgenplus3",
        "p": "constpval",
        "q": "constqval",
        "r": "constrval"
      },
      "question": "10. Given the power-series\n\\[\ncoefzero+coefone \\, variablex+coeftwo \\, variablex^{2}+\\cdots\n\\]\nin which\n\\[\ncoefgeneral=\\left(indexvar^{2}+1\\right) 3^{indexvar}\n\\]\nshow that there is a relation of the form\n\\[\ncoefgeneral+constpval \\, coefgennext+constqval \\, coefgenplus2+constrval \\, coefgenplus3=0\n\\]\nin which \\( constpval, constqval, constrval \\) are constants independent of \\( indexvar \\). Find these constants and the sum of the power-series.",
      "solution": "First Solution. The desired relation is\n\\[\n\\begin{array}{c}\n\\left(indexvar^{2}+1\\right) 3^{indexvar}+constpval\\left((indexvar+1)^{2}+1\\right) 3^{indexvar+1} \\\\\n+constqval\\left((indexvar+2)^{2}+1\\right) 3^{indexvar+2}+constrval\\left((indexvar+3)^{2}+1\\right) 3^{indexvar+3}=0\n\\end{array}\n\\]\nwhich is equivalent to\n\\[\n\\begin{array}{c}\nindexvar^{2}(1+3 constpval+9 constqval+27 constrval)+indexvar(6 constpval+36 constqval+162 constrval) \\\\\n\\quad+(1+6 constpval+45 constqval+270 constrval)=0\n\\end{array}\n\\]\n\nEquation (1) holds for all \\( indexvar \\) if and only if\n\\[\n\\begin{array}{r}\n1+3 constpval+9 constqval+27 constrval=0 \\\\\nconstpval+6 constqval+27 constrval=0 \\\\\n1+6 constpval+45 constqval+270 constrval=0\n\\end{array}\n\\]\n\nThese linear equations have the solution \\( constpval=-1, constqval=\\frac{1}{3}, constrval=-\\frac{1}{27} \\), so\n\\[\ncoefgeneral-coefgennext+\\frac{1}{3} coefgenplus2-\\frac{1}{27} coefgenplus3=0\n\\]\n\nLet \\( seriesfunc(variablex)=coefzero+coefone \\, variablex+coeftwo \\, variablex^{2}+\\cdots \\). Proceeding formally, we have\n\\[\n\\begin{array}{rlrl}\nvariablex^{3} seriesfunc(variablex) & = & coefzero \\, variablex^{3}+coefone \\, variablex^{4}+\\cdots+coefgenminus3 \\, variablex^{indexvar}+\\cdots \\\\\nconstpval \\, variablex^{2} seriesfunc(variablex) & = & constpval \\, coefzero \\, variablex^{2}+constpval \\, coefone \\, variablex^{3}+constpval \\, coeftwo \\, variablex^{4}+\\cdots+constpval \\, coefgenminus2 \\, variablex^{indexvar}+\\cdots \\\\\nconstqval \\, variablex \\, seriesfunc(variablex) & = & constqval \\, coefzero \\, variablex+constqval \\, coefone \\, variablex^{2}+constqval \\, coeftwo \\, variablex^{3}+constqval \\, coefthree \\, variablex^{4}+\\cdots+constqval \\, coefgenminus1 \\, variablex^{indexvar}+\\cdots \\\\\nconstrval \\, seriesfunc(variablex) & = & constrval \\, coefzero+constrval \\, coefone \\, variablex+constrval \\, coeftwo \\, variablex^{2}+constrval \\, coefthree \\, variablex^{3}+constrval \\, coeffour \\, variablex^{4}+\\cdots+constrval \\, coefgeneral \\, variablex^{indexvar}+\\cdots\n\\end{array}\n\\]\n\nWhen we sum these we get\n\\[\nseriesfunc(variablex)\\left[variablex^{3}+constpval \\, variablex^{2}+constqval \\, variablex+constrval\\right]=\\left(constpval \\, coefzero+constqval \\, coefone+constrval \\, coeftwo\\right) variablex^{2}+\\left(constqval \\, coefzero+constrval \\, coefone\\right) variablex+constrval \\, coefzero\n\\]\n\nMultiplying through by -27 , we obtain\n\\[\nseriesfunc(variablex)\\left[1-9 variablex+27 variablex^{2}-27 variablex^{3}\\right]=1-3 variablex+18 variablex^{2}\n\\]\nand therefore\n\\[\nseriesfunc(variablex)=\\frac{1-3 variablex+18 variablex^{2}}{(1-3 variablex)^{3}}\n\\]\n\nUsing the ratio test we conclude that the series converges for \\( |variablex|<\\frac{1}{3} \\); hence the formal manipulations above are valid for these values of \\( variablex \\).\n\nSecond Solution. Let \\( altcoefgeneral=coefgeneral / 3^{indexvar}=indexvar^{2}+1 \\). Then\n\\[\n\\begin{aligned}\n\\Delta altcoefgeneral & =altcoefgennext-altcoefgeneral=2 indexvar+1, \\quad \\Delta^{2} altcoefgeneral=altcoefgenplus2-2 altcoefgennext+altcoefgeneral=2 \\\\\n\\Delta^{3} altcoefgeneral & =altcoefgenplus3-3 altcoefgenplus2+3 altcoefgennext-altcoefgeneral=0\n\\end{aligned}\n\\]\n\nSo\n\\[\n\\frac{coefgenplus3}{3^{indexvar+3}}-\\frac{3 coefgenplus2}{3^{indexvar+2}}+\\frac{3 coefgennext}{3^{indexvar+1}}-\\frac{coefgeneral}{3^{indexvar}}=0\n\\]\nwhence\n\\[\ncoefgeneral-coefgennext+(1 / 3) coefgenplus2-(1 / 27) coefgenplus3=0\n\\]\n\nSince\n\\[\nindexvar^{2}+1=(indexvar+1)(indexvar+2)-3(indexvar+1)+2\n\\]\nwe have\n\\[\n\\begin{aligned}\n\\Sigma altcoefgeneral \\, variabley^{indexvar} & =\\Sigma(indexvar+1)(indexvar+2) \\, variabley^{indexvar}-3 \\Sigma(indexvar+1) \\, variabley^{indexvar}+2 \\Sigma \\, variabley^{indexvar} \\\\\n& =\\frac{2}{(1-variabley)^{3}}-\\frac{3}{(1-variabley)^{2}}+\\frac{2}{1-variabley} \\\\\n& =\\frac{1-variabley+2 variabley^{2}}{(1-variabley)^{3}}\n\\end{aligned}\n\\]\nprovided \\( |variabley|<1 \\). Replace \\( variabley \\) by \\( 3 \\, variablex \\).\n\\[\n\\Sigma coefgeneral \\, variablex^{indexvar}=\\frac{1-3 variablex+18 variablex^{2}}{(1-3 variablex)^{3}}\n\\]\nprovided \\( |variablex|<\\frac{1}{3} \\).\nRemark. We could assume that the problem is concerned with the ring of formal power series. In that case, \\( (1-3 variablex) \\) has an inverse in the ring and our result is that\n\\[\n\\sum coefgeneral \\, variablex^{indexvar}=\\left(1-3 variablex+18 variablex^{2}\\right)(1-3 variablex)^{-3}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "y": "driftwood",
        "n": "fireplace",
        "S": "breadcrumb",
        "a_0": "raincloud",
        "a_1": "snowflake",
        "a_2": "thunderstorm",
        "a_3": "hillcrest",
        "a_4": "stargazer",
        "a_n": "treeladder",
        "a_n+1": "riverstone",
        "a_n+2": "canyonpass",
        "a_n+3": "meadowlark",
        "a_n-3": "fernshadow",
        "a_n-2": "cliffhaven",
        "a_n-1": "brookfield",
        "b_n": "lanternfly",
        "b_n+1": "emberglow",
        "b_n+2": "nightshade",
        "b_n+3": "silverpine",
        "p": "sunflower",
        "q": "moonstone",
        "r": "oceanbreeze"
      },
      "question": "10. Given the power-series\n\\[\nraincloud+snowflake sandstone+thunderstorm sandstone^{2}+\\cdots\n\\]\nin which\n\\[\ntreeladder=\\left(fireplace^{2}+1\\right) 3^{fireplace}\n\\]\nshow that there is a relation of the form\n\\[\ntreeladder+sunflower riverstone+moonstone canyonpass+oceanbreeze meadowlark=0\n\\]\nin which \\( sunflower, moonstone, oceanbreeze \\) are constants independent of \\( fireplace \\). Find these constants and the sum of the power-series.",
      "solution": "First Solution. The desired relation is\n\\[\n\\begin{array}{c}\n\\left(fireplace^{2}+1\\right) 3^{fireplace}+sunflower\\left((fireplace+1)^{2}+1\\right) 3^{fireplace+1} \\\\\n+moonstone\\left((fireplace+2)^{2}+1\\right) 3^{fireplace+2}+oceanbreeze\\left((fireplace+3)^{2}+1\\right) 3^{fireplace+3}=0\n\\end{array}\n\\]\nwhich is equivalent to\n\\[\n\\begin{array}{c}\nfireplace^{2}(1+3 sunflower+9 moonstone+27 oceanbreeze)+fireplace(6 sunflower+36 moonstone+162 oceanbreeze) \\\\\n\\quad+(1+6 sunflower+45 moonstone+270 oceanbreeze)=0\n\\end{array}\n\\]\n\nEquation (1) holds for all \\( fireplace \\) if and only if\n\\[\n\\begin{array}{r}\n1+3 sunflower+9 moonstone+27 oceanbreeze=0 \\\\\nsunflower+6 moonstone+27 oceanbreeze=0 \\\\\n1+6 sunflower+45 moonstone+270 oceanbreeze=0\n\\end{array}\n\\]\n\nThese linear equations have the solution \\( sunflower=-1, moonstone=\\frac{1}{3}, oceanbreeze=-\\frac{1}{27} \\), so\n\\[\ntreeladder-riverstone+\\frac{1}{3} canyonpass-\\frac{1}{27} meadowlark=0\n\\]\n\nLet \\( breadcrumb(sandstone)=raincloud+snowflake sandstone+thunderstorm sandstone^{2}+\\cdots \\). Proceeding formally, we have\n\\[\n\\begin{array}{rlrl}\nsandstone^{3} breadcrumb(sandstone) & = & raincloud sandstone^{3}+snowflake sandstone^{4}+\\cdots+fernshadow sandstone^{fireplace}+\\cdots \\\\\nsunflower sandstone^{2} breadcrumb(sandstone) & = & sunflower raincloud sandstone^{2}+sunflower snowflake sandstone^{3}+sunflower thunderstorm sandstone^{4}+\\cdots+sunflower cliffhaven sandstone^{fireplace}+\\cdots \\\\\nmoonstone sandstone breadcrumb(sandstone) & = & moonstone raincloud sandstone+moonstone snowflake sandstone^{2}+moonstone thunderstorm sandstone^{3}+moonstone hillcrest sandstone^{4}+\\cdots+moonstone brookfield sandstone^{fireplace}+\\cdots \\\\\noceanbreeze breadcrumb(sandstone) & =oceanbreeze raincloud+oceanbreeze snowflake sandstone+oceanbreeze thunderstorm sandstone^{2}+oceanbreeze hillcrest sandstone^{3}+oceanbreeze stargazer sandstone^{4}+\\cdots+oceanbreeze treeladder sandstone^{fireplace}+\\cdots\n\\end{array}\n\\]\n\nWhen we sum these we get\n\\[\nbreadcrumb(sandstone)\\left[sandstone^{3}+sunflower sandstone^{2}+moonstone sandstone+oceanbreeze\\right]=\\left(sunflower raincloud+moonstone snowflake+oceanbreeze thunderstorm\\right) sandstone^{2}+\\left(moonstone raincloud+oceanbreeze snowflake\\right) sandstone+oceanbreeze raincloud\n\\]\n\nMultiplying through by -27 , we obtain\n\\[\nbreadcrumb(sandstone)\\left[1-9 sandstone+27 sandstone^{2}-27 sandstone^{3}\\right]=1-3 sandstone+18 sandstone^{2}\n\\]\nand therefore\n\\[\nbreadcrumb(sandstone)=\\frac{1-3 sandstone+18 sandstone^{2}}{(1-3 sandstone)^{3}}\n\\]\n\nUsing the ratio test we conclude that the series converges for \\( |sandstone|<\\frac{1}{3} \\); hence the formal manipulations above are valid for these values of \\( sandstone \\).\n\nSecond Solution. Let \\( lanternfly=\\frac{treeladder}{3^{fireplace}}=fireplace^{2}+1 \\). Then\n\\[\n\\begin{aligned}\n\\Delta lanternfly & =emberglow-lanternfly=2 fireplace+1, \\quad \\Delta^{2} lanternfly=nightshade-2 emberglow+lanternfly=2 \\\\\n\\Delta^{3} lanternfly & =silverpine-3 nightshade+3 emberglow-lanternfly=0\n\\end{aligned}\n\\]\n\nSo\n\\[\n\\frac{meadowlark}{3^{fireplace+3}}-\\frac{3 canyonpass}{3^{fireplace+2}}+\\frac{3 riverstone}{3^{fireplace+1}}-\\frac{treeladder}{3^{fireplace}}=0\n\\]\nwhence\n\\[\ntreeladder-riverstone+(1 / 3) canyonpass-(1 / 27) meadowlark=0\n\\]\n\nSince\n\\[\nfireplace^{2}+1=(fireplace+1)(fireplace+2)-3(fireplace+1)+2\n\\]\nwe have\n\\[\n\\begin{aligned}\n\\Sigma lanternfly driftwood^{fireplace} & =\\Sigma(fireplace+1)(fireplace+2) driftwood^{fireplace}-3 \\Sigma(fireplace+1) driftwood^{fireplace}+2 \\Sigma driftwood^{fireplace} \\\\\n& =\\frac{2}{(1-driftwood)^{3}}-\\frac{3}{(1-driftwood)^{2}}+\\frac{2}{1-driftwood} \\\\\n& =\\frac{1-driftwood+2 driftwood^{2}}{(1-driftwood)^{3}}\n\\end{aligned}\n\\]\nprovided \\( |driftwood|<1 \\). Replace \\( driftwood \\) by \\( 3 sandstone \\).\n\\[\n\\Sigma treeladder sandstone^{fireplace}=\\frac{1-3 sandstone+18 sandstone^{2}}{(1-3 sandstone)^{3}}\n\\]\nprovided \\( |sandstone|<\\frac{1}{3} \\).\nRemark. We could assume that the problem is concerned with the ring of formal power series. In that case, \\( (1-3 sandstone) \\) has an inverse in the ring and our result is that\n\\[\n\\sum treeladder sandstone^{fireplace}=\\left(1-3 sandstone+18 sandstone^{2}\\right)(1-3 sandstone)^{-3}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constant",
        "y": "steadyval",
        "n": "endpoint",
        "S": "difference",
        "a_0": "uncertainzero",
        "a_1": "uncertainone",
        "a_2": "uncertaintwo",
        "a_3": "uncertainthree",
        "a_4": "uncertainfour",
        "a_n": "uncertainindex",
        "a_n+1": "uncertainplusone",
        "a_n+2": "uncertainplustwo",
        "a_n+3": "uncertainplusthree",
        "a_n-3": "uncertainminusthree",
        "a_n-2": "uncertainminustwo",
        "a_n-1": "uncertainminusone",
        "b_n": "volatileindex",
        "b_n+1": "volatileplusone",
        "b_n+2": "volatileplustwo",
        "b_n+3": "volatileplusthree",
        "p": "staticcoef",
        "q": "stillcoef",
        "r": "rigidcoef"
      },
      "question": "10. Given the power-series\n\\[\nuncertainzero+uncertainone constant+uncertaintwo constant^{2}+\\cdots\n\\]\nin which\n\\[\nuncertainindex=\\left(endpoint^{2}+1\\right) 3^{endpoint}\n\\]\nshow that there is a relation of the form\n\\[\nuncertainindex+staticcoef uncertainplusone+stillcoef uncertainplustwo+rigidcoef uncertainplusthree=0\n\\]\nin which \\( staticcoef, stillcoef, rigidcoef \\) are constants independent of \\( endpoint \\). Find these constants and the sum of the power-series.",
      "solution": "First Solution. The desired relation is\n\\[\n\\begin{array}{c}\n\\left(endpoint^{2}+1\\right) 3^{endpoint}+staticcoef\\left((endpoint+1)^{2}+1\\right) 3^{endpoint+1} \\\\\n+stillcoef\\left((endpoint+2)^{2}+1\\right) 3^{endpoint+2}+rigidcoef\\left((endpoint+3)^{2}+1\\right) 3^{endpoint+3}=0\n\\end{array}\n\\]\nwhich is equivalent to\n\\[\n\\begin{array}{c}\nendpoint^{2}(1+3 staticcoef+9 stillcoef+27 rigidcoef)+endpoint(6 staticcoef+36 stillcoef+162 rigidcoef) \\\\\n\\quad+(1+6 staticcoef+45 stillcoef+270 rigidcoef)=0\n\\end{array}\n\\]\n\nEquation (1) holds for all \\( endpoint \\) if and only if\n\\[\n\\begin{array}{r}\n1+3 staticcoef+9 stillcoef+27 rigidcoef=0 \\\\\nstaticcoef+6 stillcoef+27 rigidcoef=0 \\\\\n1+6 staticcoef+45 stillcoef+270 rigidcoef=0\n\\end{array}\n\\]\n\nThese linear equations have the solution \\( staticcoef=-1, stillcoef=\\frac{1}{3}, rigidcoef=-\\frac{1}{27} \\), so\n\\[\nuncertainindex-uncertainplusone+\\frac{1}{3} uncertainplustwo-\\frac{1}{27} uncertainplusthree=0\n\\]\n\nLet \\( difference(constant)=uncertainzero+uncertainone constant+uncertaintwo constant^{2}+\\cdots \\). Proceeding formally, we have\n\\[\n\\begin{array}{rlrl}\nconstant^{3} difference(constant) & = & uncertainzero constant^{3}+uncertainone constant^{4}+\\cdots+uncertainminusthree constant^{endpoint}+\\cdots \\\\\nstaticcoef constant^{2} difference(constant) & = & staticcoef uncertainzero constant^{2}+staticcoef uncertainone constant^{3}+staticcoef uncertaintwo constant^{4}+\\cdots+staticcoef uncertainminustwo constant^{endpoint}+\\cdots \\\\\nstillcoef constant difference(constant) & = & stillcoef uncertainzero constant+stillcoef uncertainone constant^{2}+stillcoef uncertaintwo constant^{3}+stillcoef uncertainthree constant^{4}+\\cdots+stillcoef uncertainminusone constant^{endpoint}+\\cdots \\\\\nrigidcoef difference(constant) & = & rigidcoef uncertainzero+rigidcoef uncertainone constant+rigidcoef uncertaintwo constant^{2}+rigidcoef uncertainthree constant^{3}+rigidcoef uncertainfour constant^{4}+\\cdots+rigidcoef uncertainindex constant^{endpoint}+\\cdots\n\\end{array}\n\\]\n\nWhen we sum these we get\n\\[\ndifference(constant)\\left[constant^{3}+staticcoef constant^{2}+stillcoef constant+rigidcoef\\right]=\\left(staticcoef uncertainzero+stillcoef uncertainone+rigidcoef uncertaintwo\\right) constant^{2}+\\left(stillcoef uncertainzero+rigidcoef uncertainone\\right) constant+rigidcoef uncertainzero\n\\]\n\nMultiplying through by -27 , we obtain\n\\[\ndifference(constant)\\left[1-9 constant+27 constant^{2}-27 constant^{3}\\right]=1-3 constant+18 constant^{2}\n\\]\nand therefore\n\\[\ndifference(constant)=\\frac{1-3 constant+18 constant^{2}}{(1-3 constant)^{3}}\n\\]\n\nUsing the ratio test we conclude that the series converges for \\( |constant|<\\frac{1}{3} \\); hence the formal manipulations above are valid for these values of \\( constant \\).\n\nSecond Solution. Let \\( volatileindex=uncertainindex / 3^{endpoint}=endpoint^{2}+1 \\). Then\n\\[\n\\begin{aligned}\n\\Delta volatileindex & =volatileplusone-volatileindex=2 endpoint+1, \\quad \\Delta^{2} volatileindex=volatileplustwo-2 volatileplusone+volatileindex=2 \\\\\n\\Delta^{3} volatileindex & =volatileplusthree-3 volatileplustwo+3 volatileplusone-volatileindex=0\n\\end{aligned}\n\\]\n\nSo\n\\[\n\\frac{uncertainplusthree}{3^{endpoint+3}}-\\frac{3 uncertainplustwo}{3^{endpoint+2}}+\\frac{3 uncertainplusone}{3^{endpoint+1}}-\\frac{uncertainindex}{3^{endpoint}}=0\n\\]\nwhence\n\\[\nuncertainindex-uncertainplusone+(1 / 3) uncertainplustwo-\\left(1 / 27\\right) uncertainplusthree=0\n\\]\n\nSince\n\\[\nendpoint^{2}+1=(endpoint+1)(endpoint+2)-3(endpoint+1)+2\n\\]\nwe have\n\\[\n\\begin{aligned}\n\\Sigma volatileindex steadyval^{endpoint} & =\\Sigma(endpoint+1)(endpoint+2) steadyval^{endpoint}-3 \\Sigma(endpoint+1) steadyval^{endpoint}+2 \\Sigma steadyval^{endpoint} \\\\\n& =\\frac{2}{(1-steadyval)^{3}}-\\frac{3}{(1-steadyval)^{2}}+\\frac{2}{1-steadyval} \\\\\n& =\\frac{1-steadyval+2 steadyval^{2}}{(1-steadyval)^{3}}\n\\end{aligned}\n\\]\nprovided \\( |steadyval|<1 \\). Replace \\( steadyval \\) by \\( 3 constant \\).\n\\[\n\\Sigma uncertainindex constant^{endpoint}=\\frac{1-3 constant+18 constant^{2}}{(1-3 constant)^{3}}\n\\]\nprovided \\( |constant|<\\frac{1}{3} \\).\nRemark. We could assume that the problem is concerned with the ring of formal power series. In that case, \\( (1-3 constant) \\) has an inverse in the ring and our result is that\n\\[\n\\sum uncertainindex constant^{endpoint}=\\left(1-3 constant+18 constant^{2}\\right)(1-3 constant)^{-3}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "n": "fpldsear",
        "S": "mnbvcxqe",
        "a_0": "lakdjfgh",
        "a_1": "qwerpoiu",
        "a_2": "zmxncvas",
        "a_3": "bnmhytre",
        "a_4": "plokijuh",
        "a_n": "asdfghjk",
        "a_n+1": "ghjklasd",
        "a_n+2": "uioplkjh",
        "a_n+3": "xcvbnmas",
        "a_n-3": "wertyuii",
        "a_n-2": "sdfghjkk",
        "a_n-1": "cvbnmqqw",
        "b_n": "lkjhgfdx",
        "b_n+1": "poiuytre",
        "b_n+2": "mnbvcxzq",
        "b_n+3": "zxcvbnmm",
        "p": "rtyuioop",
        "q": "fghjklzx",
        "r": "vcxzlkjh"
      },
      "question": "10. Given the power-series\n\\[\nlakdjfgh+qwerpoiu qzxwvtnp+zmxncvas qzxwvtnp^{2}+\\cdots\n\\]\nin which\n\\[\nasdfghjk=\\left(fpldsear^{2}+1\\right) 3^{fpldsear}\n\\]\nshow that there is a relation of the form\n\\[\nasdfghjk+rtyuioop ghjklasd+fghjklzx uioplkjh+vcxzlkjh xcvbnmas=0\n\\]\nin which \\( rtyuioop, fghjklzx, vcxzlkjh \\) are constants independent of \\( fpldsear \\). Find these constants and the sum of the power-series.",
      "solution": "First Solution. The desired relation is\n\\[\n\\begin{array}{c}\n\\left(fpldsear^{2}+1\\right) 3^{fpldsear}+rtyuioop\\left((fpldsear+1)^{2}+1\\right) 3^{fpldsear+1} \\\\\n+fghjklzx\\left((fpldsear+2)^{2}+1\\right) 3^{fpldsear+2}+vcxzlkjh\\left((fpldsear+3)^{2}+1\\right) 3^{fpldsear+3}=0\n\\end{array}\n\\]\nwhich is equivalent to\n\\[\n\\begin{array}{c}\nfpldsear^{2}(1+3 rtyuioop+9 fghjklzx+27 vcxzlkjh)+fpldsear(6 rtyuioop+36 fghjklzx+162 vcxzlkjh) \\\\\n\\quad+(1+6 rtyuioop+45 fghjklzx+270 vcxzlkjh)=0\n\\end{array}\n\\]\n\nEquation (1) holds for all \\( fpldsear \\) if and only if\n\\[\n\\begin{array}{r}\n1+3 rtyuioop+9 fghjklzx+27 vcxzlkjh=0 \\\\\nrtyuioop+6 fghjklzx+27 vcxzlkjh=0 \\\\\n1+6 rtyuioop+45 fghjklzx+270 vcxzlkjh=0\n\\end{array}\n\\]\n\nThese linear equations have the solution \\( rtyuioop=-1, fghjklzx=\\frac{1}{3}, vcxzlkjh=-\\frac{1}{27} \\), so\n\\[\nasdfghjk-ghjklasd+\\frac{1}{3} uioplkjh-\\frac{1}{27} xcvbnmas=0\n\\]\n\nLet \\( mnbvcxqe(qzxwvtnp)=lakdjfgh+qwerpoiu qzxwvtnp+zmxncvas qzxwvtnp^{2}+\\cdots \\). Proceeding formally, we have\n\\[\n\\begin{array}{rlrl}\nqzxwvtnp^{3} mnbvcxqe(qzxwvtnp) & = & lakdjfgh qzxwvtnp^{3}+qwerpoiu qzxwvtnp^{4}+\\cdots+wertyuii qzxwvtnp^{fpldsear}+\\cdots \\\\\nrtyuioop qzxwvtnp^{2} mnbvcxqe(qzxwvtnp) & = & rtyuioop lakdjfgh qzxwvtnp^{2}+rtyuioop qwerpoiu qzxwvtnp^{3}+rtyuioop zmxncvas qzxwvtnp^{4}+\\cdots+rtyuioop sdfghjkk qzxwvtnp^{fpldsear}+\\cdots \\\\\nfghjklzx qzxwvtnp mnbvcxqe(qzxwvtnp) & = & fghjklzx lakdjfgh qzxwvtnp+fghjklzx qwerpoiu qzxwvtnp^{2}+fghjklzx zmxncvas qzxwvtnp^{3}+fghjklzx bnmhytre qzxwvtnp^{4}+\\cdots+fghjklzx cvbnmqqw qzxwvtnp^{fpldsear}+\\cdots \\\\\nvcxzlkjh mnbvcxqe(qzxwvtnp) & = & vcxzlkjh lakdjfgh+vcxzlkjh qwerpoiu qzxwvtnp+vcxzlkjh zmxncvas qzxwvtnp^{2}+vcxzlkjh bnmhytre qzxwvtnp^{3}+vcxzlkjh plokijuh qzxwvtnp^{4}+\\cdots+vcxzlkjh asdfghjk qzxwvtnp^{fpldsear}+\\cdots\n\\end{array}\n\\]\n\nWhen we sum these we get\n\\[\nmnbvcxqe(qzxwvtnp)\\left[qzxwvtnp^{3}+rtyuioop qzxwvtnp^{2}+fghjklzx qzxwvtnp+vcxzlkjh\\right]=\\left(rtyuioop lakdjfgh+fghjklzx qwerpoiu+vcxzlkjh zmxncvas\\right) qzxwvtnp^{2}+\\left(fghjklzx lakdjfgh+vcxzlkjh qwerpoiu\\right) qzxwvtnp+vcxzlkjh lakdjfgh\n\\]\n\nMultiplying through by -27 , we obtain\n\\[\nmnbvcxqe(qzxwvtnp)\\left[1-9 qzxwvtnp+27 qzxwvtnp^{2}-27 qzxwvtnp^{3}\\right]=1-3 qzxwvtnp+18 qzxwvtnp^{2}\n\\]\nand therefore\n\\[\nmnbvcxqe(qzxwvtnp)=\\frac{1-3 qzxwvtnp+18 qzxwvtnp^{2}}{(1-3 qzxwvtnp)^{3}}\n\\]\n\nUsing the ratio test we conclude that the series converges for \\( |qzxwvtnp|<\\frac{1}{3} \\); hence the formal manipulations above are valid for these values of \\( qzxwvtnp \\).\n\nSecond Solution. Let \\( lkjhgfdx=asdfghjk / 3^{fpldsear}=fpldsear^{2}+1 \\). Then\n\\[\n\\begin{aligned}\n\\Delta lkjhgfdx & =poiuytre-lkjhgfdx=2 fpldsear+1, \\quad \\Delta^{2} lkjhgfdx=mnbvcxzq-2 poiuytre+lkjhgfdx=2 \\\\\n\\Delta^{3} lkjhgfdx & =zxcvbnmm-3 mnbvcxzq+3 poiuytre-lkjhgfdx=0\n\\end{aligned}\n\\]\n\nSo\n\\[\n\\frac{xcvbnmas}{3^{fpldsear+3}}-\\frac{3 uioplkjh}{3^{fpldsear+2}}+\\frac{3 ghjklasd}{3^{fpldsear+1}}-\\frac{asdfghjk}{3^{fpldsear}}=0\n\\]\nwhence\n\\[\nasdfghjk-ghjklasd+(1 / 3) uioplkjh-(1 / 27) xcvbnmas=0\n\\]\n\nSince\n\\[\nfpldsear^{2}+1=(fpldsear+1)(fpldsear+2)-3(fpldsear+1)+2\n\\]\nwe have\n\\[\n\\begin{aligned}\n\\Sigma lkjhgfdx hjgrksla^{fpldsear} & =\\Sigma(fpldsear+1)(fpldsear+2) hjgrksla^{fpldsear}-3 \\Sigma(fpldsear+1) hjgrksla^{fpldsear}+2 \\Sigma hjgrksla^{fpldsear} \\\\\n& =\\frac{2}{(1-hjgrksla)^{3}}-\\frac{3}{(1-hjgrksla)^{2}}+\\frac{2}{1-hjgrksla} \\\\\n& =\\frac{1-hjgrksla+2 hjgrksla^{2}}{(1-hjgrksla)^{3}}\n\\end{aligned}\n\\]\nprovided \\( |hjgrksla|<1 \\). Replace \\( hjgrksla \\) by \\( 3 qzxwvtnp \\).\n\\[\n\\Sigma asdfghjk qzxwvtnp^{fpldsear}=\\frac{1-3 qzxwvtnp+18 qzxwvtnp^{2}}{(1-3 qzxwvtnp)^{3}}\n\\]\nprovided \\( |qzxwvtnp|<\\frac{1}{3} \\).\nRemark. We could assume that the problem is concerned with the ring of formal power series. In that case, \\( (1-3 qzxwvtnp) \\) has an inverse in the ring and our result is that\n\\[\n\\sum asdfghjk qzxwvtnp^{fpldsear}=\\left(1-3 qzxwvtnp+18 qzxwvtnp^{2}\\right)(1-3 qzxwvtnp)^{-3}\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\\[\na_{n}=n(n-1)(n-2)(n-3)\\,2^{n}\\;+\\;\\bigl(n^{2}+1\\bigr)(-3)^{n}\\qquad(n\\ge 0)\n\\]\nand let its ordinary generating function be  \n\\[\nS(x)=\\sum_{n=0}^{\\infty}a_{n}\\,x^{n}.\n\\]\n\n1.\\;(Linear annihilator)\\;Prove that there exist constants  \n\\[\nc_{0},c_{1},\\dots ,c_{8}\\in\\mathbb{C},\\qquad c_{0}\\neq 0,\n\\]\nindependent of $n$, such that for every $n\\ge 0$\n\\[\nc_{0}a_{n+8}+c_{1}a_{n+7}+\\dots +c_{7}a_{n+1}+c_{8}a_{n}=0.\n\\]\n\n2.\\;(Minimal order \\& exact coefficients)  \na)\\;Show that no relation of order $\\le 7$ is possible, so the relation in (1) is of minimal length.  \nb)\\;Determine the unique integer coefficients (up to a common non–zero factor) and write the recurrence explicitly.\n\n3.\\;(Closed rational form)  \nProve directly from the definition of $a_{n}$ that $S(x)$ is annihilated by the polynomial $(1-2x)^{5}(1+3x)^{3}$ and hence\n\\[\nS(x)=\\frac{P(x)}{(1-2x)^{5}(1+3x)^{3}},\n\\qquad\\deg P\\le 7.\n\\]\nCompute the polynomial $P(x)$ explicitly.\n\n4.\\;(Analytic information)  \nLocate every singularity of $S(x)$ in $\\mathbb{C}$ and determine the exact radius of convergence of the power series.\n\n5.\\;(Numerical evaluation)  \nEvaluate the convergent series\n\\[\n\\sum_{n=0}^{\\infty}\\frac{a_{n}}{6^{n}}\n\\]\nin closed form.\n\n\\vspace{.5em}",
      "solution": "1.\\;Existence of an order–$8$ linear relation  \n\nLet $E$ be the forward–shift operator, $(E\\,f)_{n}=f_{n+1}$.\n\n(i)\\;$n(n-1)(n-2)(n-3)\\,2^{n}$ is a degree–$4$ polynomial in $n$ multiplied by $2^{n}$, hence it is annihilated by $(E-2)^{5}$ (the $5^{\\text{th}}$ difference of a degree–$4$ polynomial evaluated at $2^{n}$ vanishes).\n\n(ii)\\;$(n^{2}+1)(-3)^{n}$ is a degree–$2$ polynomial times $(-3)^{n}$, hence it is annihilated by $(E+3)^{3}$.\n\nSince $(E-2)^{5}\\,(E+3)^{3}$ annihilates each summand, it annihilates their sum $a_{n}$.  \nExpanding the product gives a monic polynomial of degree $8$ in $E$, so an order–$8$ homogeneous linear recurrence with constant coefficients exists.\n\n2.\\;The minimal recurrence and its coefficients  \n\na)\\;Minimality  \n\nThe factors $E-2$ and $E+3$ are coprime as polynomials in $E$.  \nFor any linear operator $L(E)$ with constant coefficients that annihilates $a_{n}$, we have\n\\[\nL(E)=U(E)\\,(E-2)^{5}(E+3)^{3},\n\\]\nbecause $(E-2)^{5}(E+3)^{3}$ is a greatest common right–divisor of the two annihilators obtained in 1(i) and 1(ii).  \n(One may invoke Bézout’s identity for polynomials: since $\\gcd(E-2,E+3)=1$ there exist polynomials $A,B$ with $A(E-2)+B(E+3)=1$, and hence any common multiple of $(E-2)^{5}$ and $(E+3)^{3}$ is divisible by their product.)  \nTherefore $\\deg L\\ge 5+3=8$, so no recurrence of order $\\le 7$ can exist; the order $8$ found above is minimal.\n\nb)\\;Explicit coefficients  \n\nThe recurrence coefficients are those of the polynomial\n\\[\n(E-2)^{5}(E+3)^{3}\\;=\\;\\sum_{k=0}^{8}c_{k}\\,E^{8-k},\n\\]\nread from highest to lowest degree.  Equivalently, they are the coefficients of\n\\[\n(1-2x)^{5}(1+3x)^{3}=\\sum_{k=0}^{8}c_{k}\\,x^{k}.\n\\]\nCompute  \n\\[\n\\begin{aligned}\n(1-2x)^{5}&=1-10x+40x^{2}-80x^{3}+80x^{4}-32x^{5},\\\\\n(1+3x)^{3}&=1+9x+27x^{2}+27x^{3},\n\\end{aligned}\n\\]\nthen multiply:\n\\[\n\\bigl(1-2x\\bigr)^{5}\\bigl(1+3x\\bigr)^{3}\n=1- x-23x^{2}+37x^{3}+170x^{4}-392x^{5}-288x^{6}+1296x^{7}-864x^{8}.\n\\]\nHence, for every $n\\ge 0$,\n\\[\n\\boxed{%\na_{n+8}-a_{n+7}-23a_{n+6}+37a_{n+5}+170a_{n+4}\n-392a_{n+3}-288a_{n+2}+1296a_{n+1}-864a_{n}=0}.\n\\]\n\n3.\\;Closed rational form of $S(x)$  \n\nSeparate $a_{n}$ into two parts:\n\n(i)\\;$\\displaystyle \\sum_{n\\ge 0}n(n-1)(n-2)(n-3)\\,2^{n}x^{n}$\n\n\\[\n=\\frac{4!\\,(2x)^{4}}{(1-2x)^{5}}\n=\\frac{384\\,x^{4}}{(1-2x)^{5}}.\n\\]\n\n(ii)\\;$\\displaystyle \\sum_{n\\ge 0}(n^{2}+1)(-3)^{n}x^{n}$  \n\nWrite $n^{2}+1=n(n-1)+n+1$ and use the standard formulae for the sums of powers:\n\n\\[\n\\begin{aligned}\n\\sum_{n\\ge 0}n(n-1)(-3x)^{n}&=\\frac{2(-3x)^{2}}{(1+3x)^{3}},\\\\\n\\sum_{n\\ge 0}n(-3x)^{n}&=\\frac{-3x}{(1+3x)^{2}},\\\\\n\\sum_{n\\ge 0}(-3x)^{n}&=\\frac{1}{1+3x}.\n\\end{aligned}\n\\]\nAdding gives\n\\[\n\\sum_{n\\ge 0}(n^{2}+1)(-3x)^{n}\n=\\frac{1+3x+18x^{2}}{(1+3x)^{3}}.\n\\]\n\nAdd (i) and (ii) over the common denominator $(1-2x)^{5}(1+3x)^{3}$:\n\n\\[\n\\begin{aligned}\nP(x)&=384x^{4}(1+3x)^{3}+(1+3x+18x^{2})(1-2x)^{5}\\\\[2pt]\n&=1-7x+28x^{2}-140x^{3}+944x^{4}\\\\\n&\\quad +2224x^{5}+11712x^{6}+9792x^{7}.\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\boxed{%\nS(x)=\\dfrac{1-7x+28x^{2}-140x^{3}+944x^{4}+2224x^{5}+11712x^{6}+9792x^{7}}\n{(1-2x)^{5}(1+3x)^{3}} }.\n\\]\n\n4.\\;Singularities and radius of convergence  \n\n$S(x)$ is a rational function.  Its only singularities are the poles coming from the denominator:\n\n\\[\nx=\\frac12\\quad(\\text{order }5),\\qquad x=-\\frac13\\quad(\\text{order }3).\n\\]\n\nHence the power series about $x=0$ converges in the largest open disc avoiding these poles, namely\n\n\\[\nR=\\min\\Bigl\\{\\bigl|\\,\\tfrac12\\bigr|,\\bigl|-\\tfrac13\\bigr|\\Bigr\\}=\\frac13.\n\\]\n\nTherefore the series converges for $|x|<\\dfrac13$ and diverges for $|x|>\\dfrac13$.  \n(Only two points on the circle $|x|=\\dfrac13$ are singular; analytic continuation is possible through every other point of the circle, so the circle is the boundary of convergence but not a natural boundary in the classical sense.)\n\n5.\\;Numerical evaluation at $x=\\dfrac16$  \n\nBecause $\\dfrac16<R$, substitution into the closed form is legitimate.\n\n\\[\n\\begin{aligned}\n(1-2x)^{5}&=\\Bigl(1-\\frac13\\Bigr)^{5}=\\Bigl(\\frac23\\Bigr)^{5}=\\frac{32}{243},\\\\\n(1+3x)^{3}&=\\Bigl(1+\\frac12\\Bigr)^{3}=\\Bigl(\\frac32\\Bigr)^{3}=\\frac{27}{8}.\n\\end{aligned}\n\\]\n\nCompute the two partial sums exactly:\n\n(i)\\;First part  \n\\[\n\\frac{384x^{4}}{(1-2x)^{5}}\n=\\frac{384\\left(\\dfrac16\\right)^{4}}{\\dfrac{32}{243}}\n=\\frac{8/27}{32/243}=\\frac{9}{4}.\n\\]\n\n(ii)\\;Second part  \n\\[\n\\frac{1+3x+18x^{2}}{(1+3x)^{3}}\n=\\frac{1+\\dfrac12+\\dfrac12}{27/8}\n=\\frac{2}{27/8}=\\frac{16}{27}.\n\\]\n\nAdding yields\n\\[\n\\boxed{\\displaystyle \\sum_{n=0}^{\\infty}\\frac{a_{n}}{6^{n}}\n=S\\!\\bigl(\\tfrac16\\bigr)=\\frac94+\\frac{16}{27}=\\frac{307}{108}.}\n\\]\n\n\\vspace{.5em}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.359386",
        "was_fixed": false,
        "difficulty_analysis": "• Higher degree & mixed bases The first summand involves a degree-4 polynomial while the second uses a degree-2 polynomial and the opposite sign base (−3).  Their combination forces an 8-term recurrence, double the length of the original variant.\n\n• Multiple distinct characteristic roots Because 2 and −3 are coprime roots, students must factor and combine two high-multiplicity annihilators, recognise coprimality and minimality, and handle much larger computations.\n\n• Algebraic heft Expanding (1−2x)⁵(1+3x)³ and constructing P(x) entails manipulating polynomials up to degree 8, far more cumbersome than the quadratic–cubic case of the original.\n\n• Analytic component Locating all complex singularities and proving the radius of convergence injects complex-analysis reasoning absent from the original.\n\n• Numerical evaluation A non-trivial rational value (307/108) must be distilled from the rational function, demanding careful arithmetic and checking.\n\nOverall the variant intertwines high-order difference operators, generating-function calculus, polynomial algebra, and analytic considerations, substantially deepening both the conceptual and computational load compared with the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\na_{n}=n(n-1)(n-2)(n-3)\\,2^{n}\\;+\\;\\bigl(n^{2}+1\\bigr)(-3)^{n}\\qquad(n\\ge 0)\n\\]\nand let its ordinary generating function be  \n\\[\nS(x)=\\sum_{n=0}^{\\infty}a_{n}\\,x^{n}.\n\\]\n\n1.\\;(Linear annihilator)\\;Prove that there exist constants  \n\\[\nc_{0},c_{1},\\dots ,c_{8}\\in\\mathbb{C},\\qquad c_{0}\\neq 0,\n\\]\nindependent of $n$, such that for every $n\\ge 0$\n\\[\nc_{0}a_{n+8}+c_{1}a_{n+7}+\\dots +c_{7}a_{n+1}+c_{8}a_{n}=0.\n\\]\n\n2.\\;(Minimal order \\& exact coefficients)  \na)\\;Show that no relation of order $\\le 7$ is possible, so the relation in (1) is of minimal length.  \nb)\\;Determine the unique integer coefficients (up to a common non–zero factor) and write the recurrence explicitly.\n\n3.\\;(Closed rational form)  \nProve directly from the definition of $a_{n}$ that $S(x)$ is annihilated by the polynomial $(1-2x)^{5}(1+3x)^{3}$ and hence\n\\[\nS(x)=\\frac{P(x)}{(1-2x)^{5}(1+3x)^{3}},\n\\qquad\\deg P\\le 7.\n\\]\nCompute the polynomial $P(x)$ explicitly.\n\n4.\\;(Analytic information)  \nLocate every singularity of $S(x)$ in $\\mathbb{C}$ and determine the exact radius of convergence of the power series.\n\n5.\\;(Numerical evaluation)  \nEvaluate the convergent series\n\\[\n\\sum_{n=0}^{\\infty}\\frac{a_{n}}{6^{n}}\n\\]\nin closed form.\n\n\\vspace{.5em}",
      "solution": "1.\\;Existence of an order–$8$ linear relation  \n\nLet $E$ be the forward–shift operator, $(E\\,f)_{n}=f_{n+1}$.\n\n(i)\\;$n(n-1)(n-2)(n-3)\\,2^{n}$ is a degree–$4$ polynomial in $n$ multiplied by $2^{n}$, hence it is annihilated by $(E-2)^{5}$ (the $5^{\\text{th}}$ difference of a degree–$4$ polynomial evaluated at $2^{n}$ vanishes).\n\n(ii)\\;$(n^{2}+1)(-3)^{n}$ is a degree–$2$ polynomial times $(-3)^{n}$, hence it is annihilated by $(E+3)^{3}$.\n\nSince $(E-2)^{5}\\,(E+3)^{3}$ annihilates each summand, it annihilates their sum $a_{n}$.  \nExpanding the product gives a monic polynomial of degree $8$ in $E$, so an order–$8$ homogeneous linear recurrence with constant coefficients exists.\n\n2.\\;The minimal recurrence and its coefficients  \n\na)\\;Minimality  \n\nThe factors $E-2$ and $E+3$ are coprime as polynomials in $E$.  \nFor any linear operator $L(E)$ with constant coefficients that annihilates $a_{n}$, we have\n\\[\nL(E)=U(E)\\,(E-2)^{5}(E+3)^{3},\n\\]\nbecause $(E-2)^{5}(E+3)^{3}$ is a greatest common right–divisor of the two annihilators obtained in 1(i) and 1(ii).  \n(One may invoke Bézout’s identity for polynomials: since $\\gcd(E-2,E+3)=1$ there exist polynomials $A,B$ with $A(E-2)+B(E+3)=1$, and hence any common multiple of $(E-2)^{5}$ and $(E+3)^{3}$ is divisible by their product.)  \nTherefore $\\deg L\\ge 5+3=8$, so no recurrence of order $\\le 7$ can exist; the order $8$ found above is minimal.\n\nb)\\;Explicit coefficients  \n\nThe recurrence coefficients are those of the polynomial\n\\[\n(E-2)^{5}(E+3)^{3}\\;=\\;\\sum_{k=0}^{8}c_{k}\\,E^{8-k},\n\\]\nread from highest to lowest degree.  Equivalently, they are the coefficients of\n\\[\n(1-2x)^{5}(1+3x)^{3}=\\sum_{k=0}^{8}c_{k}\\,x^{k}.\n\\]\nCompute  \n\\[\n\\begin{aligned}\n(1-2x)^{5}&=1-10x+40x^{2}-80x^{3}+80x^{4}-32x^{5},\\\\\n(1+3x)^{3}&=1+9x+27x^{2}+27x^{3},\n\\end{aligned}\n\\]\nthen multiply:\n\\[\n\\bigl(1-2x\\bigr)^{5}\\bigl(1+3x\\bigr)^{3}\n=1- x-23x^{2}+37x^{3}+170x^{4}-392x^{5}-288x^{6}+1296x^{7}-864x^{8}.\n\\]\nHence, for every $n\\ge 0$,\n\\[\n\\boxed{%\na_{n+8}-a_{n+7}-23a_{n+6}+37a_{n+5}+170a_{n+4}\n-392a_{n+3}-288a_{n+2}+1296a_{n+1}-864a_{n}=0}.\n\\]\n\n3.\\;Closed rational form of $S(x)$  \n\nSeparate $a_{n}$ into two parts:\n\n(i)\\;$\\displaystyle \\sum_{n\\ge 0}n(n-1)(n-2)(n-3)\\,2^{n}x^{n}$\n\n\\[\n=\\frac{4!\\,(2x)^{4}}{(1-2x)^{5}}\n=\\frac{384\\,x^{4}}{(1-2x)^{5}}.\n\\]\n\n(ii)\\;$\\displaystyle \\sum_{n\\ge 0}(n^{2}+1)(-3)^{n}x^{n}$  \n\nWrite $n^{2}+1=n(n-1)+n+1$ and use the standard formulae for the sums of powers:\n\n\\[\n\\begin{aligned}\n\\sum_{n\\ge 0}n(n-1)(-3x)^{n}&=\\frac{2(-3x)^{2}}{(1+3x)^{3}},\\\\\n\\sum_{n\\ge 0}n(-3x)^{n}&=\\frac{-3x}{(1+3x)^{2}},\\\\\n\\sum_{n\\ge 0}(-3x)^{n}&=\\frac{1}{1+3x}.\n\\end{aligned}\n\\]\nAdding gives\n\\[\n\\sum_{n\\ge 0}(n^{2}+1)(-3x)^{n}\n=\\frac{1+3x+18x^{2}}{(1+3x)^{3}}.\n\\]\n\nAdd (i) and (ii) over the common denominator $(1-2x)^{5}(1+3x)^{3}$:\n\n\\[\n\\begin{aligned}\nP(x)&=384x^{4}(1+3x)^{3}+(1+3x+18x^{2})(1-2x)^{5}\\\\[2pt]\n&=1-7x+28x^{2}-140x^{3}+944x^{4}\\\\\n&\\quad +2224x^{5}+11712x^{6}+9792x^{7}.\n\\end{aligned}\n\\]\n\nThus\n\\[\n\\boxed{%\nS(x)=\\dfrac{1-7x+28x^{2}-140x^{3}+944x^{4}+2224x^{5}+11712x^{6}+9792x^{7}}\n{(1-2x)^{5}(1+3x)^{3}} }.\n\\]\n\n4.\\;Singularities and radius of convergence  \n\n$S(x)$ is a rational function.  Its only singularities are the poles coming from the denominator:\n\n\\[\nx=\\frac12\\quad(\\text{order }5),\\qquad x=-\\frac13\\quad(\\text{order }3).\n\\]\n\nHence the power series about $x=0$ converges in the largest open disc avoiding these poles, namely\n\n\\[\nR=\\min\\Bigl\\{\\bigl|\\,\\tfrac12\\bigr|,\\bigl|-\\tfrac13\\bigr|\\Bigr\\}=\\frac13.\n\\]\n\nTherefore the series converges for $|x|<\\dfrac13$ and diverges for $|x|>\\dfrac13$.  \n(Only two points on the circle $|x|=\\dfrac13$ are singular; analytic continuation is possible through every other point of the circle, so the circle is the boundary of convergence but not a natural boundary in the classical sense.)\n\n5.\\;Numerical evaluation at $x=\\dfrac16$  \n\nBecause $\\dfrac16<R$, substitution into the closed form is legitimate.\n\n\\[\n\\begin{aligned}\n(1-2x)^{5}&=\\Bigl(1-\\frac13\\Bigr)^{5}=\\Bigl(\\frac23\\Bigr)^{5}=\\frac{32}{243},\\\\\n(1+3x)^{3}&=\\Bigl(1+\\frac12\\Bigr)^{3}=\\Bigl(\\frac32\\Bigr)^{3}=\\frac{27}{8}.\n\\end{aligned}\n\\]\n\nCompute the two partial sums exactly:\n\n(i)\\;First part  \n\\[\n\\frac{384x^{4}}{(1-2x)^{5}}\n=\\frac{384\\left(\\dfrac16\\right)^{4}}{\\dfrac{32}{243}}\n=\\frac{8/27}{32/243}=\\frac{9}{4}.\n\\]\n\n(ii)\\;Second part  \n\\[\n\\frac{1+3x+18x^{2}}{(1+3x)^{3}}\n=\\frac{1+\\dfrac12+\\dfrac12}{27/8}\n=\\frac{2}{27/8}=\\frac{16}{27}.\n\\]\n\nAdding yields\n\\[\n\\boxed{\\displaystyle \\sum_{n=0}^{\\infty}\\frac{a_{n}}{6^{n}}\n=S\\!\\bigl(\\tfrac16\\bigr)=\\frac94+\\frac{16}{27}=\\frac{307}{108}.}\n\\]\n\n\\vspace{.5em}",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.313226",
        "was_fixed": false,
        "difficulty_analysis": "• Higher degree & mixed bases The first summand involves a degree-4 polynomial while the second uses a degree-2 polynomial and the opposite sign base (−3).  Their combination forces an 8-term recurrence, double the length of the original variant.\n\n• Multiple distinct characteristic roots Because 2 and −3 are coprime roots, students must factor and combine two high-multiplicity annihilators, recognise coprimality and minimality, and handle much larger computations.\n\n• Algebraic heft Expanding (1−2x)⁵(1+3x)³ and constructing P(x) entails manipulating polynomials up to degree 8, far more cumbersome than the quadratic–cubic case of the original.\n\n• Analytic component Locating all complex singularities and proving the radius of convergence injects complex-analysis reasoning absent from the original.\n\n• Numerical evaluation A non-trivial rational value (307/108) must be distilled from the rational function, demanding careful arithmetic and checking.\n\nOverall the variant intertwines high-order difference operators, generating-function calculus, polynomial algebra, and analytic considerations, substantially deepening both the conceptual and computational load compared with the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}