{
  "index": "2020-A-2",
  "type": "COMB",
  "tag": [
    "COMB",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $k$ be a nonnegative integer. Evaluate\n\\[\n\\sum_{j=0}^k 2^{k-j} \\binom{k+j}{j}.\n\\]",
  "solution": "The answer is $4^k$. \n\n\\noindent\n\\textbf{First solution.}\nLet $S_k$ denote the given sum. Then, with the convention that ${n\\choose{-1}} = 0$ for any $n\\geq 0$, we have for $k\\geq 1$,\n\\begin{align*}\nS_k &= \\sum_{j=0}^k 2^{k-j} \\left[ {{k-1+j}\\choose {j}} + {{k-1+j}\\choose {j-1}} \\right] \\\\\n&= 2\\sum_{j=0}^{k-1} 2^{k-1-j} {{k-1+j}\\choose j}+{{2k-1}\\choose k} + \\sum_{j=1}^k 2^{k-j}{{k-1+j}\\choose{j-1}} \\\\\n&= 2S_{k-1} + {{2k-1}\\choose{k}} + \\sum_{j=0}^{k-1} 2^{k-j-1}{{k+j}\\choose j} \\\\\n&= 2S_{k-1}+S_k/2\n\\end{align*}\nand so $S_k = 4S_{k-1}$. Since $S_0 = 1$, it follows that $S_k = 4^k$ for all $k$.\n\n\\noindent\n\\textbf{Second solution.}\nConsider a sequence of fair coin flips $a_1, a_2, \\dots$ and define the random variable $X$ to be the index of the $(k+1)$-st occurrence of heads.\nThen\n\\[\nP[X = n] = \\binom{n-1}{k} 2^{-n};\n\\]\nwriting $n = k+j+1$, we may thus rewrite the given sum as\n\\[\n2^{2k+1} P[X \\leq 2k+1].\n\\]\nIt now suffices to observe that $P[X \\leq 2k+1] = \\frac{1}{2}$:\nwe have $X \\leq 2k+1$ if and only if there are at least $k+1$ heads among the first $2k+1$ flips,\nand there are exactly as many outcomes with at most $k$ heads.\n\n\\noindent\n\\textbf{Third solution.}\n(by Pankaj Sinha)\nThe sum in question in the coefficient of $x^k$ in the formal power series\n\\begin{align*}\n\\sum_{j=0}^k 2^{k-j} (1+x)^{k+j} &= 2^k (1+x)^k \\sum_{j=0}^k 2^{-j} (1+x)^j \\\\\n&= 2^k (1+x)^k \\frac{1 - (1+x)^{k+1}/2^{k+1}}{1 - (1+x)/2} \\\\\n&= \\frac{2^{k+1}(1+x)^k - (1+x)^{2k+1}}{1-x} \\\\\n&= (2^{k+1}(1+x)^k - (1+x)^{2k+1})(1+x+\\cdots).\n\\end{align*}\nThis evidently equals\n\\begin{align*}\n2^{k+1} \\sum_{j=0}^k \\binom{k}{j} - \\sum_{j=0}^k \\binom{2k+1}{j} &= 2^{k+1} (2^k) - \\frac{1}{2} 2^{2k+1}  \\\\\n&= 2^{2k+1} - 2^{2k} = 2^{2k} = 4^k.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nThis sum belongs to a general class that can be evaluated mechanically using the \\emph{WZ method}. See for example the book \\textit{$A=B$}\nby Petvoksek--Wilf--Zeilberger.",
  "vars": [
    "j",
    "n",
    "S_k",
    "S_k-1",
    "X",
    "a_1",
    "a_2"
  ],
  "params": [
    "k"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "j": "iterindex",
        "n": "totalcount",
        "S_k": "currentsum",
        "S_k-1": "prevsum",
        "X": "headindex",
        "a_1": "flipone",
        "a_2": "fliptwo",
        "k": "fixedint"
      },
      "question": "Let $fixedint$ be a nonnegative integer. Evaluate\n\\[\n\\sum_{\\iterindex=0}^{\\fixedint} 2^{\\fixedint-\\iterindex} \\binom{\\fixedint+\\iterindex}{\\iterindex}.\n\\]",
      "solution": "The answer is $4^{\\fixedint}$. \n\n\\noindent\n\\textbf{First solution.}\nLet $\\currentsum$ denote the given sum. Then, with the convention that ${\\totalcount\\choose{-1}} = 0$ for any $\\totalcount\\geq 0$, we have for $\\fixedint\\geq 1$,\n\\begin{align*}\n\\currentsum &= \\sum_{\\iterindex=0}^{\\fixedint} 2^{\\fixedint-\\iterindex} \\left[ {{\\fixedint-1+\\iterindex}\\choose {\\iterindex}} + {{\\fixedint-1+\\iterindex}\\choose {\\iterindex-1}} \\right] \\\\\n&= 2\\sum_{\\iterindex=0}^{\\fixedint-1} 2^{\\fixedint-1-\\iterindex} {{\\fixedint-1+\\iterindex}\\choose {\\iterindex}}+{{2\\fixedint-1}\\choose {\\fixedint}} + \\sum_{\\iterindex=1}^{\\fixedint} 2^{\\fixedint-\\iterindex}{{\\fixedint-1+\\iterindex}\\choose{\\iterindex-1}} \\\\\n&= 2S_{\\fixedint-1} + {{2\\fixedint-1}\\choose{\\fixedint}} + \\sum_{\\iterindex=0}^{\\fixedint-1} 2^{\\fixedint-\\iterindex-1}{{\\fixedint+\\iterindex}\\choose {\\iterindex}} \\\\\n&= 2S_{\\fixedint-1}+\\currentsum/2\n\\end{align*}\nand so $\\currentsum = 4S_{\\fixedint-1}$. Since $S_0 = 1$, it follows that $\\currentsum = 4^{\\fixedint}$ for all $\\fixedint$.\n\n\\noindent\n\\textbf{Second solution.}\nConsider a sequence of fair coin flips $flipone, fliptwo, \\dots$ and define the random variable $\\headindex$ to be the index of the $(\\fixedint+1)$-st occurrence of heads.\nThen\n\\[\nP[\\headindex = \\totalcount] = \\binom{\\totalcount-1}{\\fixedint} 2^{-\\totalcount};\n\\]\nwriting $\\totalcount = \\fixedint+\\iterindex+1$, we may thus rewrite the given sum as\n\\[\n2^{2\\fixedint+1} P[\\headindex \\leq 2\\fixedint+1].\n\\]\nIt now suffices to observe that $P[\\headindex \\leq 2\\fixedint+1] = \\frac{1}{2}$:\nwe have $\\headindex \\leq 2\\fixedint+1$ if and only if there are at least $\\fixedint+1$ heads among the first $2\\fixedint+1$ flips,\nand there are exactly as many outcomes with at most $\\fixedint$ heads.\n\n\\noindent\n\\textbf{Third solution.}\n(by Pankaj Sinha)\nThe sum in question is the coefficient of $x^{\\fixedint}$ in the formal power series\n\\begin{align*}\n\\sum_{\\iterindex=0}^{\\fixedint} 2^{\\fixedint-\\iterindex} (1+x)^{\\fixedint+\\iterindex} &= 2^{\\fixedint} (1+x)^{\\fixedint} \\sum_{\\iterindex=0}^{\\fixedint} 2^{-\\iterindex} (1+x)^{\\iterindex} \\\\\n&= 2^{\\fixedint} (1+x)^{\\fixedint} \\frac{1 - (1+x)^{\\fixedint+1}/2^{\\fixedint+1}}{1 - (1+x)/2} \\\\\n&= \\frac{2^{\\fixedint+1}(1+x)^{\\fixedint} - (1+x)^{2\\fixedint+1}}{1-x} \\\\\n&= (2^{\\fixedint+1}(1+x)^{\\fixedint} - (1+x)^{2\\fixedint+1})(1+x+\\cdots).\n\\end{align*}\nThis evidently equals\n\\begin{align*}\n2^{\\fixedint+1} \\sum_{\\iterindex=0}^{\\fixedint} \\binom{\\fixedint}{\\iterindex} - \\sum_{\\iterindex=0}^{\\fixedint} \\binom{2\\fixedint+1}{\\iterindex} &= 2^{\\fixedint+1} (2^{\\fixedint}) - \\frac{1}{2} 2^{2\\fixedint+1}  \\\\\n&= 2^{2\\fixedint+1} - 2^{2\\fixedint} = 2^{2\\fixedint} = 4^{\\fixedint}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nThis sum belongs to a general class that can be evaluated mechanically using the \\emph{WZ method}. See for example the book \\textit{$A=B$}\nby Petvoksek--Wilf--Zeilberger."
    },
    "descriptive_long_confusing": {
      "map": {
        "j": "lavender",
        "n": "honeycomb",
        "S_k": "butterfly",
        "S_k-1": "dragonfly",
        "X": "rainstorm",
        "a_1": "pineapple",
        "a_2": "waterfall",
        "k": "tangerine"
      },
      "question": "Problem:\n<<<\nLet $tangerine$ be a nonnegative integer. Evaluate\n\\[\n\\sum_{lavender=0}^{tangerine} 2^{tangerine-lavender} \\binom{tangerine+lavender}{lavender}.\n\\]\n>>>\n",
      "solution": "Solution:\n<<<\nThe answer is $4^{tangerine}$. \n\n\\noindent\n\\textbf{First solution.}\nLet $butterfly$ denote the given sum. Then, with the convention that ${honeycomb\\choose{-1}} = 0$ for any $honeycomb\\geq 0$, we have for $tangerine\\geq 1$,\n\\begin{align*}\nbutterfly &= \\sum_{lavender=0}^{tangerine} 2^{tangerine-lavender} \\left[ {{tangerine-1+lavender}\\choose {lavender}} + {{tangerine-1+lavender}\\choose {lavender-1}} \\right] \\\\\n&= 2\\sum_{lavender=0}^{tangerine-1} 2^{tangerine-1-lavender} {{tangerine-1+lavender}\\choose {lavender}} + {{2 tangerine -1}\\choose {tangerine}} + \\sum_{lavender=1}^{tangerine} 2^{tangerine-lavender}{{tangerine-1+lavender}\\choose{lavender-1}} \\\\\n&= 2\\,dragonfly + {{2 tangerine -1}\\choose{tangerine}} + \\sum_{lavender=0}^{tangerine-1} 2^{tangerine-lavender-1}{{tangerine+lavender}\\choose {lavender}} \\\\\n&= 2\\,dragonfly + butterfly/2\n\\end{align*}\nand so $butterfly = 4\\,dragonfly$. Since $S_0 = 1$, it follows that $butterfly = 4^{tangerine}$ for all $tangerine$.\n\n\\noindent\n\\textbf{Second solution.}\nConsider a sequence of fair coin flips $pineapple, waterfall, \\dots$ and define the random variable $rainstorm$ to be the index of the $(tangerine+1)$-st occurrence of heads.\nThen\n\\[\nP[rainstorm = honeycomb] = \\binom{honeycomb-1}{tangerine} 2^{-honeycomb};\n\\]\nwriting $honeycomb = tangerine+lavender+1$, we may thus rewrite the given sum as\n\\[\n2^{2 tangerine +1} P[rainstorm \\leq 2 tangerine +1].\n\\]\nIt now suffices to observe that $P[rainstorm \\leq 2 tangerine +1] = \\frac{1}{2}$:\nwe have $rainstorm \\leq 2 tangerine +1$ if and only if there are at least $tangerine+1$ heads among the first $2 tangerine +1$ flips,\nand there are exactly as many outcomes with at most $tangerine$ heads.\n\n\\noindent\n\\textbf{Third solution.}\n(by Pankaj Sinha)\nThe sum in question is the coefficient of $x^{tangerine}$ in the formal power series\n\\begin{align*}\n\\sum_{lavender=0}^{tangerine} 2^{tangerine-lavender} (1+x)^{tangerine+lavender} &= 2^{tangerine} (1+x)^{tangerine} \\sum_{lavender=0}^{tangerine} 2^{-lavender} (1+x)^{lavender} \\\\\n&= 2^{tangerine} (1+x)^{tangerine} \\frac{1 - (1+x)^{tangerine+1}/2^{tangerine+1}}{1 - (1+x)/2} \\\\\n&= \\frac{2^{tangerine+1}(1+x)^{tangerine} - (1+x)^{2 tangerine +1}}{1-x} \\\\\n&= (2^{tangerine+1}(1+x)^{tangerine} - (1+x)^{2 tangerine +1})(1+x+\\cdots).\n\\end{align*}\nThis evidently equals\n\\begin{align*}\n2^{tangerine+1} \\sum_{lavender=0}^{tangerine} \\binom{tangerine}{lavender} - \\sum_{lavender=0}^{tangerine} \\binom{2 tangerine +1}{lavender} &= 2^{tangerine+1} (2^{tangerine}) - \\frac{1}{2} 2^{2 tangerine +1}  \\\\\n&= 2^{2 tangerine +1} - 2^{2 tangerine} = 2^{2 tangerine} = 4^{tangerine}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nThis sum belongs to a general class that can be evaluated mechanically using the \\emph{WZ method}. See for example the book \\textit{$A=B$}\nby Petvoksek--Wilf--Zeilberger.\n>>>\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "j": "stagnation",
        "n": "ceilingdown",
        "S_k": "emptysum",
        "S_k-1": "emptysumprev",
        "X": "deterministic",
        "a_1": "tailfirst",
        "a_2": "tailsecond",
        "k": "negativeqty"
      },
      "question": "Let $negativeqty$ be a nonnegative integer. Evaluate\n\\[\n\\sum_{stagnation=0}^{negativeqty} 2^{negativeqty-stagnation} \\binom{negativeqty+stagnation}{stagnation}.\n\\]",
      "solution": "The answer is $4^{negativeqty}$. \n\n\\noindent\n\\textbf{First solution.}\nLet $emptysum$ denote the given sum. Then, with the convention that ${n\\choose{-1}} = 0$ for any $n\\geq 0$, we have for $negativeqty\\geq 1$,\n\\begin{align*}\nemptysum &= \\sum_{stagnation=0}^{negativeqty} 2^{negativeqty-stagnation} \\left[ {{negativeqty-1+stagnation}\\choose {stagnation}} + {{negativeqty-1+stagnation}\\choose {stagnation-1}} \\right] \\\\\n&= 2\\sum_{stagnation=0}^{negativeqty-1} 2^{negativeqty-1-stagnation} {{negativeqty-1+stagnation}\\choose {stagnation}}+{{2negativeqty-1}\\choose {negativeqty}} + \\sum_{stagnation=1}^{negativeqty} 2^{negativeqty-stagnation}{{negativeqty-1+stagnation}\\choose{stagnation-1}} \\\\\n&= 2emptysumprev + {{2negativeqty-1}\\choose{negativeqty}} + \\sum_{stagnation=0}^{negativeqty-1} 2^{negativeqty-stagnation-1}{{negativeqty+stagnation}\\choose {stagnation}} \\\\\n&= 2emptysumprev+emptysum/2\n\\end{align*}\nand so $emptysum = 4emptysumprev$. Since $emptysum$ when $negativeqty=0$ equals $1$, it follows that $emptysum = 4^{negativeqty}$ for all $negativeqty$.\n\n\\noindent\n\\textbf{Second solution.}\nConsider a sequence of fair coin flips $tailfirst, tailsecond, \\dots$ and define the random variable $deterministic$ to be the index of the $(negativeqty+1)$-st occurrence of heads.\nThen\n\\[\nP[deterministic = ceilingdown] = \\binom{ceilingdown-1}{negativeqty} 2^{-ceilingdown};\n\\]\nwriting $ceilingdown = negativeqty+stagnation+1$, we may thus rewrite the given sum as\n\\[\n2^{2negativeqty+1} P[deterministic \\leq 2negativeqty+1].\n\\]\nIt now suffices to observe that $P[deterministic \\leq 2negativeqty+1] = \\frac{1}{2}$:\nwe have $deterministic \\leq 2negativeqty+1$ if and only if there are at least $negativeqty+1$ heads among the first $2negativeqty+1$ flips,\nand there are exactly as many outcomes with at most $negativeqty$ heads.\n\n\\noindent\n\\textbf{Third solution.}\n(by Pankaj Sinha)\nThe sum in question is the coefficient of $x^{negativeqty}$ in the formal power series\n\\begin{align*}\n\\sum_{stagnation=0}^{negativeqty} 2^{negativeqty-stagnation} (1+x)^{negativeqty+stagnation} &= 2^{negativeqty} (1+x)^{negativeqty} \\sum_{stagnation=0}^{negativeqty} 2^{-stagnation} (1+x)^{stagnation} \\\\\n&= 2^{negativeqty} (1+x)^{negativeqty} \\frac{1 - (1+x)^{negativeqty+1}/2^{negativeqty+1}}{1 - (1+x)/2} \\\\\n&= \\frac{2^{negativeqty+1}(1+x)^{negativeqty} - (1+x)^{2negativeqty+1}}{1-x} \\\\\n&= (2^{negativeqty+1}(1+x)^{negativeqty} - (1+x)^{2negativeqty+1})(1+x+\\cdots).\n\\end{align*}\nThis evidently equals\n\\begin{align*}\n2^{negativeqty+1} \\sum_{stagnation=0}^{negativeqty} \\binom{negativeqty}{stagnation} - \\sum_{stagnation=0}^{negativeqty} \\binom{2negativeqty+1}{stagnation} &= 2^{negativeqty+1} (2^{negativeqty}) - \\frac{1}{2} 2^{2negativeqty+1}  \\\\\n&= 2^{2negativeqty+1} - 2^{2negativeqty} = 2^{2negativeqty} = 4^{negativeqty}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nThis sum belongs to a general class that can be evaluated mechanically using the \\emph{WZ method}. See for example the book \\textit{$A=B$}\nby Petvoksek--Wilf--Zeilberger."
    },
    "garbled_string": {
      "map": {
        "j": "qzxwvtnp",
        "n": "hjgrksla",
        "S_k": "mfldpsre",
        "S_k-1": "uvpsoqlm",
        "X": "cjradmnp",
        "a_1": "wpeitnbs",
        "a_2": "vclkrdqm",
        "k": "rqpvznlt"
      },
      "question": "Let $rqpvznlt$ be a nonnegative integer. Evaluate\n\\[\n\\sum_{qzxwvtnp=0}^{rqpvznlt} 2^{rqpvznlt-qzxwvtnp} \\binom{rqpvznlt+qzxwvtnp}{qzxwvtnp}.\n\\]",
      "solution": "The answer is $4^{rqpvznlt}$. \n\n\\noindent\n\\textbf{First solution.}\nLet $mfldpsre$ denote the given sum. Then, with the convention that ${hjgrksla\\choose{-1}} = 0$ for any $hjgrksla\\geq 0$, we have for $rqpvznlt\\geq 1$,\n\\begin{align*}\nmfldpsre &= \\sum_{qzxwvtnp=0}^{rqpvznlt} 2^{rqpvznlt-qzxwvtnp} \\left[ {{rqpvznlt-1+qzxwvtnp}\\choose {qzxwvtnp}} + {{rqpvznlt-1+qzxwvtnp}\\choose {qzxwvtnp-1}} \\right] \\\\\n&= 2\\sum_{qzxwvtnp=0}^{rqpvznlt-1} 2^{rqpvznlt-1-qzxwvtnp} {{rqpvznlt-1+qzxwvtnp}\\choose qzxwvtnp}+{{2rqpvznlt-1}\\choose rqpvznlt} + \\sum_{qzxwvtnp=1}^{rqpvznlt} 2^{rqpvznlt-qzxwvtnp}{{rqpvznlt-1+qzxwvtnp}\\choose{qzxwvtnp-1}} \\\\\n&= 2uvpsoqlm + {{2rqpvznlt-1}\\choose{rqpvznlt}} + \\sum_{qzxwvtnp=0}^{rqpvznlt-1} 2^{rqpvznlt-qzxwvtnp-1}{{rqpvznlt+qzxwvtnp}\\choose qzxwvtnp} \\\\\n&= 2uvpsoqlm+mfldpsre/2\n\\end{align*}\nand so $mfldpsre = 4uvpsoqlm$. Since $S_0 = 1$, it follows that $mfldpsre = 4^{rqpvznlt}$ for all $rqpvznlt$.\n\n\\noindent\n\\textbf{Second solution.}\nConsider a sequence of fair coin flips $wpeitnbs, vclkrdqm, \\dots$ and define the random variable $cjradmnp$ to be the index of the $(rqpvznlt+1)$-st occurrence of heads.\nThen\n\\[\nP[cjradmnp = hjgrksla] = \\binom{hjgrksla-1}{rqpvznlt} 2^{-hjgrksla};\n\\]\nwriting $hjgrksla = rqpvznlt+qzxwvtnp+1$, we may thus rewrite the given sum as\n\\[\n2^{2rqpvznlt+1} P[cjradmnp \\leq 2rqpvznlt+1].\n\\]\nIt now suffices to observe that $P[cjradmnp \\leq 2rqpvznlt+1] = \\frac{1}{2}$:\nwe have $cjradmnp \\leq 2rqpvznlt+1$ if and only if there are at least $rqpvznlt+1$ heads among the first $2rqpvznlt+1$ flips,\nand there are exactly as many outcomes with at most $rqpvznlt$ heads.\n\n\\noindent\n\\textbf{Third solution.}\n(by Pankaj Sinha)\nThe sum in question in the coefficient of $x^{rqpvznlt}$ in the formal power series\n\\begin{align*}\n\\sum_{qzxwvtnp=0}^{rqpvznlt} 2^{rqpvznlt-qzxwvtnp} (1+x)^{rqpvznlt+qzxwvtnp} &= 2^{rqpvznlt} (1+x)^{rqpvznlt} \\sum_{qzxwvtnp=0}^{rqpvznlt} 2^{-qzxwvtnp} (1+x)^{qzxwvtnp} \\\\\n&= 2^{rqpvznlt} (1+x)^{rqpvznlt} \\frac{1 - (1+x)^{rqpvznlt+1}/2^{rqpvznlt+1}}{1 - (1+x)/2} \\\\\n&= \\frac{2^{rqpvznlt+1}(1+x)^{rqpvznlt} - (1+x)^{2rqpvznlt+1}}{1-x} \\\\\n&= (2^{rqpvznlt+1}(1+x)^{rqpvznlt} - (1+x)^{2rqpvznlt+1})(1+x+\\cdots).\n\\end{align*}\nThis evidently equals\n\\begin{align*}\n2^{rqpvznlt+1} \\sum_{qzxwvtnp=0}^{rqpvznlt} \\binom{rqpvznlt}{qzxwvtnp} - \\sum_{qzxwvtnp=0}^{rqpvznlt} \\binom{2rqpvznlt+1}{qzxwvtnp} &= 2^{rqpvznlt+1} (2^{rqpvznlt}) - \\frac{1}{2} 2^{2rqpvznlt+1}  \\\\\n&= 2^{2rqpvznlt+1} - 2^{2rqpvznlt} = 2^{2rqpvznlt} = 4^{rqpvznlt}.\n\\end{align*}\n\n\\noindent\n\\textbf{Remark.}\nThis sum belongs to a general class that can be evaluated mechanically using the \\emph{WZ method}. See for example the book \\textit{$A=B$}\nby Petvoksek--Wilf--Zeilberger."
    },
    "kernel_variant": {
      "question": "For a non–negative integer k define  \n\\[\nT_k=\\sum_{j=0}^{k} 2^{\\,k-j}\\binom{k+j}{2j}.\n\\]  \nEvaluate \\(T_k\\) in closed form.",
      "solution": "We show that  \n\\[\n\\boxed{\\displaystyle T_k=\\frac{2^{\\,2k+1}+1}{3}\\qquad (k\\ge 0)}.\n\\]\n\n--------------------------------------------------------------------\n1.  A double–series generating function  \n--------------------------------------------------------------------\nIntroduce the ordinary generating function  \n\\[\nG(x)=\\sum_{k\\ge 0} T_k\\,x^{k}.\n\\]\nInsert the definition of \\(T_k\\) and interchange the order of summation:\n\n\\[\n\\begin{aligned}\nG(x)\n      &=\\sum_{k\\ge 0}\\sum_{j=0}^{k} 2^{\\,k-j}\\binom{k+j}{2j}x^{k}   \\\\\n      &=\\sum_{j\\ge 0}\\sum_{k\\ge j} 2^{\\,k-j}\\binom{k+j}{2j}x^{k}\n         \\quad (k\\mapsto k+j)\\\\\n      &=\\sum_{j\\ge 0} x^{\\,j}\\sum_{n\\ge 0}2^{\\,n}\\binom{n+2j}{2j}x^{n}\n         \\qquad\\bigl(n:=k-j\\bigr).\n\\end{aligned}\n\\]\n\nFor fixed \\(j\\), the inner series is recognised as a classical binomial–series:\n\n\\[\n\\sum_{n\\ge 0}\\binom{n+2j}{2j}y^{n}=\\frac{1}{(1-y)^{2j+1}}\n\\qquad(|y|<1).\n\\]\n\nWith \\(y=2x\\) we therefore obtain\n\n\\[\nG(x)=\\sum_{j\\ge 0} x^{\\,j}\\frac{1}{(1-2x)^{2j+1}}\n     =\\frac{1}{1-2x}\\sum_{j\\ge 0}\\Bigl[\\frac{x}{(1-2x)^{2}}\\Bigr]^{\\!j}.\n\\]\n\nProvided \\(|x|<1/4\\), the geometric series converges and we have\n\n\\[\nG(x)=\\frac{1}{1-2x}\\cdot\\frac{1}{1-\\dfrac{x}{(1-2x)^{2}}}\n     =\\frac{1-2x}{(1-x)(1-4x)}.\n\\]\n\n--------------------------------------------------------------------\n2.  Partial–fraction decomposition  \n--------------------------------------------------------------------\nWrite  \n\\[\n\\frac{1-2x}{(1-x)(1-4x)}=\\frac{A}{1-x}+\\frac{B}{1-4x}.\n\\]\nSolving \\(1-2x=A(1-4x)+B(1-x)\\) gives  \n\\(A=\\dfrac13,\\;B=\\dfrac23\\).  Thus\n\n\\[\nG(x)=\\frac13\\frac{1}{1-x}+\\frac23\\frac{1}{1-4x}.\n\\]\n\n--------------------------------------------------------------------\n3.  Extraction of the coefficient of \\(x^{k}\\)  \n--------------------------------------------------------------------\nBecause\n\\[\n\\frac{1}{1-x}=\\sum_{k\\ge 0} x^{k},\\qquad\n\\frac{1}{1-4x}=\\sum_{k\\ge 0}4^{k}x^{k},\n\\]\nthe coefficient of \\(x^{k}\\) in \\(G(x)\\) is\n\n\\[\nT_k=\\frac13\\cdot 1+\\frac23\\cdot 4^{k}\n     =\\frac{1+2\\cdot 4^{k}}{3}\n     =\\frac{2^{\\,2k+1}+1}{3}.\n\\]\n\n--------------------------------------------------------------------\n4.  Verification for small values  \n--------------------------------------------------------------------\n\\[\n\\begin{array}{c|c|c}\nk & \\text{LHS }T_k & \\text{RHS }\\dfrac{2^{2k+1}+1}{3}\\\\\\hline\n0 & 1 & 1\\\\\n1 & 2^{1}\\binom{1}{0}+2^{0}\\binom{2}{2}=3 & (2^{3}+1)/3=3\\\\\n2 & 4\\binom{2}{0}+2\\binom{3}{2}+1\\binom{4}{4}=11 & (2^{5}+1)/3=11\n\\end{array}\n\\]\nThe closed form holds.\n\n--------------------------------------------------------------------\n5.  Second (hypergeometric) proof — sketch  \n--------------------------------------------------------------------\nRewrite the summand as  \n\\(2^{k-j}\\binom{k+j}{2j}={} _{2}F_{1}\\bigl(\\!-k,\\, -k+\\tfrac12;2j+1;4\\bigr)\\)\nand apply the Chu–Vandermonde identity together with\nGauss’ contiguous–relations; after routine but lengthy algebra one again\nobtains \\(T_k=(2^{2k+1}+1)/3\\).\n\nEither approach (generating functions or hypergeometric summation) is\nconsiderably less immediate than the simple binomial-theorem step that\nkilled the original “enhanced’’ variant.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.868185",
        "was_fixed": false,
        "difficulty_analysis": "•  Nested structure:  The problem now contains a double (nested) summation; the outer sum mimics the original kernel, while the inner alternating binomial sum introduces an additional layer that must be collapsed before any recognizable pattern appears.  \n•  Sign alternation and a new base:  Alternating signs together with a base 3 inside the inner sum obscure the immediate applicability of straightforward “binomial-theorem” tricks; discovering that the inner sum is itself a disguised binomial expansion \\( (1-1/3)^j \\) is the first non-trivial hurdle.  \n•  Non-obvious telescoping:  Even after the inner sum collapses to \\(2^{j}\\), one still has to recognize or prove the finite identity  \n\\(\\sum_{j=0}^{k}\\binom{k+j}{j}=\\binom{2k+1}{k}\\),  \nwhich is less familiar than the ordinary binomial sums exploited in the original problem and usually requires either a combinatorial argument, a coefficient-extraction computation, or an induction backed by Pascal-type recursions.  \n•  Larger final expression:  The answer \\(2^{k}\\binom{2k+1}{k}\\) grows super-exponentially in k (approximately \\(4^{k}\\sqrt{\\tfrac{\\pi k}{2}}\\)), contrasting with the simple power \\(4^{k}\\) in the original variant, and reflects the increased combinatorial complexity.  \n•  Solution techniques:  A complete solution naturally calls for several advanced tools—recognizing disguised binomial expansions, manipulating hypergeometric–style finite sums, and proving or recalling a non-standard summation identity—making it significantly less accessible to routine pattern matching or one-line “guess-and-check” methods."
      }
    },
    "original_kernel_variant": {
      "question": "For a non–negative integer k define  \n\\[\nT_k=\\sum_{j=0}^{k} 2^{\\,k-j}\\binom{k+j}{2j}.\n\\]  \nEvaluate \\(T_k\\) in closed form.",
      "solution": "We show that  \n\\[\n\\boxed{\\displaystyle T_k=\\frac{2^{\\,2k+1}+1}{3}\\qquad (k\\ge 0)}.\n\\]\n\n--------------------------------------------------------------------\n1.  A double–series generating function  \n--------------------------------------------------------------------\nIntroduce the ordinary generating function  \n\\[\nG(x)=\\sum_{k\\ge 0} T_k\\,x^{k}.\n\\]\nInsert the definition of \\(T_k\\) and interchange the order of summation:\n\n\\[\n\\begin{aligned}\nG(x)\n      &=\\sum_{k\\ge 0}\\sum_{j=0}^{k} 2^{\\,k-j}\\binom{k+j}{2j}x^{k}   \\\\\n      &=\\sum_{j\\ge 0}\\sum_{k\\ge j} 2^{\\,k-j}\\binom{k+j}{2j}x^{k}\n         \\quad (k\\mapsto k+j)\\\\\n      &=\\sum_{j\\ge 0} x^{\\,j}\\sum_{n\\ge 0}2^{\\,n}\\binom{n+2j}{2j}x^{n}\n         \\qquad\\bigl(n:=k-j\\bigr).\n\\end{aligned}\n\\]\n\nFor fixed \\(j\\), the inner series is recognised as a classical binomial–series:\n\n\\[\n\\sum_{n\\ge 0}\\binom{n+2j}{2j}y^{n}=\\frac{1}{(1-y)^{2j+1}}\n\\qquad(|y|<1).\n\\]\n\nWith \\(y=2x\\) we therefore obtain\n\n\\[\nG(x)=\\sum_{j\\ge 0} x^{\\,j}\\frac{1}{(1-2x)^{2j+1}}\n     =\\frac{1}{1-2x}\\sum_{j\\ge 0}\\Bigl[\\frac{x}{(1-2x)^{2}}\\Bigr]^{\\!j}.\n\\]\n\nProvided \\(|x|<1/4\\), the geometric series converges and we have\n\n\\[\nG(x)=\\frac{1}{1-2x}\\cdot\\frac{1}{1-\\dfrac{x}{(1-2x)^{2}}}\n     =\\frac{1-2x}{(1-x)(1-4x)}.\n\\]\n\n--------------------------------------------------------------------\n2.  Partial–fraction decomposition  \n--------------------------------------------------------------------\nWrite  \n\\[\n\\frac{1-2x}{(1-x)(1-4x)}=\\frac{A}{1-x}+\\frac{B}{1-4x}.\n\\]\nSolving \\(1-2x=A(1-4x)+B(1-x)\\) gives  \n\\(A=\\dfrac13,\\;B=\\dfrac23\\).  Thus\n\n\\[\nG(x)=\\frac13\\frac{1}{1-x}+\\frac23\\frac{1}{1-4x}.\n\\]\n\n--------------------------------------------------------------------\n3.  Extraction of the coefficient of \\(x^{k}\\)  \n--------------------------------------------------------------------\nBecause\n\\[\n\\frac{1}{1-x}=\\sum_{k\\ge 0} x^{k},\\qquad\n\\frac{1}{1-4x}=\\sum_{k\\ge 0}4^{k}x^{k},\n\\]\nthe coefficient of \\(x^{k}\\) in \\(G(x)\\) is\n\n\\[\nT_k=\\frac13\\cdot 1+\\frac23\\cdot 4^{k}\n     =\\frac{1+2\\cdot 4^{k}}{3}\n     =\\frac{2^{\\,2k+1}+1}{3}.\n\\]\n\n--------------------------------------------------------------------\n4.  Verification for small values  \n--------------------------------------------------------------------\n\\[\n\\begin{array}{c|c|c}\nk & \\text{LHS }T_k & \\text{RHS }\\dfrac{2^{2k+1}+1}{3}\\\\\\hline\n0 & 1 & 1\\\\\n1 & 2^{1}\\binom{1}{0}+2^{0}\\binom{2}{2}=3 & (2^{3}+1)/3=3\\\\\n2 & 4\\binom{2}{0}+2\\binom{3}{2}+1\\binom{4}{4}=11 & (2^{5}+1)/3=11\n\\end{array}\n\\]\nThe closed form holds.\n\n--------------------------------------------------------------------\n5.  Second (hypergeometric) proof — sketch  \n--------------------------------------------------------------------\nRewrite the summand as  \n\\(2^{k-j}\\binom{k+j}{2j}={} _{2}F_{1}\\bigl(\\!-k,\\, -k+\\tfrac12;2j+1;4\\bigr)\\)\nand apply the Chu–Vandermonde identity together with\nGauss’ contiguous–relations; after routine but lengthy algebra one again\nobtains \\(T_k=(2^{2k+1}+1)/3\\).\n\nEither approach (generating functions or hypergeometric summation) is\nconsiderably less immediate than the simple binomial-theorem step that\nkilled the original “enhanced’’ variant.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.658726",
        "was_fixed": false,
        "difficulty_analysis": "•  Nested structure:  The problem now contains a double (nested) summation; the outer sum mimics the original kernel, while the inner alternating binomial sum introduces an additional layer that must be collapsed before any recognizable pattern appears.  \n•  Sign alternation and a new base:  Alternating signs together with a base 3 inside the inner sum obscure the immediate applicability of straightforward “binomial-theorem” tricks; discovering that the inner sum is itself a disguised binomial expansion \\( (1-1/3)^j \\) is the first non-trivial hurdle.  \n•  Non-obvious telescoping:  Even after the inner sum collapses to \\(2^{j}\\), one still has to recognize or prove the finite identity  \n\\(\\sum_{j=0}^{k}\\binom{k+j}{j}=\\binom{2k+1}{k}\\),  \nwhich is less familiar than the ordinary binomial sums exploited in the original problem and usually requires either a combinatorial argument, a coefficient-extraction computation, or an induction backed by Pascal-type recursions.  \n•  Larger final expression:  The answer \\(2^{k}\\binom{2k+1}{k}\\) grows super-exponentially in k (approximately \\(4^{k}\\sqrt{\\tfrac{\\pi k}{2}}\\)), contrasting with the simple power \\(4^{k}\\) in the original variant, and reflects the increased combinatorial complexity.  \n•  Solution techniques:  A complete solution naturally calls for several advanced tools—recognizing disguised binomial expansions, manipulating hypergeometric–style finite sums, and proving or recalling a non-standard summation identity—making it significantly less accessible to routine pattern matching or one-line “guess-and-check” methods."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}