{
  "index": "1992-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "power series about $x=0$ of $(1 + x)^\\alpha$. Evaluate\n\\[\n\\int_0^1 \\left( C(-y-1) \\sum_{k=1}^{1992} \\frac{1}{y+k} \\right)\\,dy.\n\\]",
  "solution": "Solution. From the binomial theorem, we see that\n\\[\nC(\\alpha)=\\alpha(\\alpha-1) \\cdots \\frac{\\alpha-1991}{1992!},\n\\]\nso \\( C(-y-1)=(y+1) \\cdots(y+1992) / 1992 \\) !. Therefore\n\\[\nC(-y-1)\\left(\\frac{1}{y+1}+\\cdots+\\frac{1}{y+1992}\\right)=\\frac{d}{d y}\\left(\\frac{(y+1) \\cdots(y+1992)}{1992!}\\right) .\n\\]\n(The same formula for the derivative of a factored polynomial came up in 1991A3.) Hence the integral in question is\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d}{d y}\\left(\\frac{(y+1) \\cdots(y+1992)}{1992!}\\right) d y & =\\left.\\frac{(y+1) \\cdots(y+1992)}{1992!}\\right|_{0} ^{1} \\\\\n& =\\frac{1993!-1992!}{1992!}=1992\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "y",
    "k"
  ],
  "params": [
    "\\\\alpha",
    "C"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "y": "variabley",
        "k": "summandk",
        "\\alpha": "exponent",
        "C": "binomcoef"
      },
      "question": "power series about $inputvar=0$ of $(1 + inputvar)^{exponent}$. Evaluate\n\\[\n\\int_0^1 \\left( binomcoef(-variabley-1) \\sum_{summandk=1}^{1992} \\frac{1}{variabley+summandk} \\right)\\,d variabley.\n\\]\n",
      "solution": "Solution. From the binomial theorem, we see that\n\\[\nbinomcoef(exponent)=exponent(exponent-1) \\cdots \\frac{exponent-1991}{1992!},\n\\]\nso \\( binomcoef(-variabley-1)=(variabley+1) \\cdots(variabley+1992) / 1992! \\). Therefore\n\\[\nbinomcoef(-variabley-1)\\left(\\frac{1}{variabley+1}+\\cdots+\\frac{1}{variabley+1992}\\right)=\\frac{d}{d variabley}\\left(\\frac{(variabley+1) \\cdots(variabley+1992)}{1992!}\\right) .\n\\]\n(The same formula for the derivative of a factored polynomial came up in 1991A3.) Hence the integral in question is\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d}{d variabley}\\left(\\frac{(variabley+1) \\cdots(variabley+1992)}{1992!}\\right) d variabley & =\\left.\\frac{(variabley+1) \\cdots(variabley+1992)}{1992!}\\right|_{0} ^{1} \\\\\n& =\\frac{1993!-1992!}{1992!}=1992\n\\end{aligned}\n\\]\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "y": "butterfly",
        "k": "lanterns",
        "\\alpha": "breakfast",
        "C": "greenhouse"
      },
      "question": "power series about $sandstone=0$ of $(1 + sandstone)^{breakfast}$. Evaluate\n\\[\n\\int_0^1 \\left( greenhouse(-butterfly-1) \\sum_{lanterns=1}^{1992} \\frac{1}{butterfly+lanterns} \\right)\\,d butterfly.\n\\]",
      "solution": "Solution. From the binomial theorem, we see that\n\\[\ngreenhouse(breakfast)=breakfast(breakfast-1) \\cdots \\frac{breakfast-1991}{1992!},\n\\]\nso \\( greenhouse(-butterfly-1)=(butterfly+1) \\cdots(butterfly+1992) / 1992 \\) !. Therefore\n\\[\ngreenhouse(-butterfly-1)\\left(\\frac{1}{butterfly+1}+\\cdots+\\frac{1}{butterfly+1992}\\right)=\\frac{d}{d butterfly}\\left(\\frac{(butterfly+1) \\cdots(butterfly+1992)}{1992!}\\right) .\n\\]\n(The same formula for the derivative of a factored polynomial came up in 1991A3.) Hence the integral in question is\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d}{d butterfly}\\left(\\frac{(butterfly+1) \\cdots(butterfly+1992)}{1992!}\\right) d butterfly & =\\left.\\frac{(butterfly+1) \\cdots(butterfly+1992)}{1992!}\\right|_{0} ^{1} \\\\\n& =\\frac{1993!-1992!}{1992!}=1992\n\\end{aligned}\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "y": "fixedvalue",
        "k": "totalcount",
        "\\alpha": "endingcoef",
        "C": "dynamicval"
      },
      "question": "power series about $constantval=0$ of $(1 + constantval)^{endingcoef}$. Evaluate\n\\[\n\\int_0^1 \\left( dynamicval(-fixedvalue-1) \\sum_{totalcount=1}^{1992} \\frac{1}{fixedvalue+totalcount} \\right)\\,d fixedvalue.\n\\]",
      "solution": "Solution. From the binomial theorem, we see that\n\\[\ndynamicval(endingcoef)=endingcoef(endingcoef-1) \\cdots \\frac{endingcoef-1991}{1992!},\n\\]\nso \\( dynamicval(-fixedvalue-1)=(fixedvalue+1) \\cdots(fixedvalue+1992) / 1992 \\) !. Therefore\n\\[\ndynamicval(-fixedvalue-1)\\left(\\frac{1}{fixedvalue+1}+\\cdots+\\frac{1}{fixedvalue+1992}\\right)=\\frac{d}{d fixedvalue}\\left(\\frac{(fixedvalue+1) \\cdots(fixedvalue+1992)}{1992!}\\right) .\n\\]\n(The same formula for the derivative of a factored polynomial came up in 1991A3.) Hence the integral in question is\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d}{d fixedvalue}\\left(\\frac{(fixedvalue+1) \\cdots(fixedvalue+1992)}{1992!}\\right) d fixedvalue & =\\left.\\frac{(fixedvalue+1) \\cdots(fixedvalue+1992)}{1992!}\\right|_{0} ^{1} \\\\\n& =\\frac{1993!-1992!}{1992!}=1992\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "k": "pcvmtzqn",
        "\\alpha": "nbvdasle",
        "C": "lopzruea"
      },
      "question": "power series about $qzxwvtnp=0$ of $(1 + qzxwvtnp)^{nbvdasle}$. Evaluate\n\\[\n\\int_0^1 \\left( lopzruea(-hjgrksla-1) \\sum_{pcvmtzqn=1}^{1992} \\frac{1}{hjgrksla+pcvmtzqn} \\right)\\,d hjgrksla.\n\\]",
      "solution": "Solution. From the binomial theorem, we see that\n\\[\nlopzruea(nbvdasle)=nbvdasle(nbvdasle-1) \\cdots \\frac{nbvdasle-1991}{1992!},\n\\]\nso \\( lopzruea(-hjgrksla-1)=(hjgrksla+1) \\cdots(hjgrksla+1992) / 1992 \\) !. Therefore\n\\[\nlopzruea(-hjgrksla-1)\\left(\\frac{1}{hjgrksla+1}+\\cdots+\\frac{1}{hjgrksla+1992}\\right)=\\frac{d}{d hjgrksla}\\left(\\frac{(hjgrksla+1) \\cdots(hjgrksla+1992)}{1992!}\\right) .\n\\]\n(The same formula for the derivative of a factored polynomial came up in 1991A3.) Hence the integral in question is\n\\[\n\\begin{aligned}\n\\int_{0}^{1} \\frac{d}{d hjgrksla}\\left(\\frac{(hjgrksla+1) \\cdots(hjgrksla+1992)}{1992!}\\right) d hjgrksla & =\\left.\\frac{(hjgrksla+1) \\cdots(hjgrksla+1992)}{1992!}\\right|_{0} ^{1} \\\\\n& =\\frac{1993!-1992!}{1992!}=1992\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let N be a positive integer (later we will specialise to N = 2024).  \nFor α∈ℝ let  \n  C_N(α) be the coefficient of x^N in the Taylor expansion of (1+x)^α about x=0, i.e.  \n\n  C_N(α)=α(α−1)⋯(α−N+1)/N!.\n\nFor y>−1 put  \n\n  S_N(y)=∑_{1≤i<j≤N} 1/[(y+i)(y+j)].\n\nEvaluate the definite integral  \n\n  I_N = ∫_0^1 C_N(−y−1)·S_N(y) dy\n\nin closed form for general N, and then give the explicit value when N = 2024.",
      "solution": "Step 1.  Write the coefficient as a factored polynomial.  \nFor all integers N≥1,\n\n C_N(−y−1)= (−1)^N (y+1)(y+2)⋯(y+N)/N!\n      = (−1)^N P_N(y)/N!,  where P_N(y)=∏_{k=1}^N (y+k).\n\nStep 2.  Express S_N(y) through derivatives of P_N.  \nBecause P_N′(y)=P_N(y)∑_{k=1}^N 1/(y+k), we have  \n\n P_N″(y)=P_N(y) [(∑_{k=1}^N 1/(y+k))^2 − ∑_{k=1}^N 1/(y+k)^2]\n     = 2 P_N(y) S_N(y).\n\nThus\n\n S_N(y)=P_N″(y)/(2P_N(y)).                     (1)\n\nStep 3.  Insert (1) into the integral.  \nUsing C_N(−y−1)=(−1)^N P_N(y)/N!,\n\n C_N(−y−1)·S_N(y)= (−1)^N P_N″(y)/(2N!).\n\nHence  \n\n I_N = ∫_0^1 (−1)^N P_N″(y)/(2N!) dy\n    = (−1)^N [P_N′(y)]_0^1 /(2N!).    (2)\n\nStep 4.  Evaluate P_N′ at y=0 and y=1.  \nWe again use P_N′(y)=P_N(y)∑_{k=1}^N 1/(y+k).\n\n•  y=0: P_N(0)=N! ⇒ P_N′(0)=N!·H_N,  \n where H_N=∑_{k=1}^N 1/k is the N-th harmonic number.\n\n•  y=1: P_N(1)=(N+1)! and ∑_{k=1}^N 1/(1+k)=H_{N+1}−1,  \n so P_N′(1)=(N+1)! (H_{N+1}−1).\n\nInsert these into (2):\n\n I_N = (−1)^N[(N+1)! (H_{N+1}−1)−N! H_N]/(2N!)\n    = (−1)^N[(N+1)(H_{N+1}−1)−H_N]/2.     (3)\n\nStep 5.  Remove H_{N+1}.  \nSince H_{N+1}=H_N+1/(N+1),\n\n (N+1)(H_{N+1}−1)−H_N\n = (N+1)(H_N−1)+1−H_N\n = N H_N−N.\n\nSubstituting in (3) yields the compact formula\n\n I_N = (−1)^N · N/2 · (H_N − 1).              (4)\n\nStep 6.  Specialise to N = 2024.  \nBecause 2024 is even, (−1)^2024 = +1, so\n\n I_2024 = 1012·(H_{2024} − 1),\n\nwhere H_{2024}=∑_{k=1}^{2024} 1/k.  This is a rational number with denominator lcm{1,2,…,2024}; its explicit fractional form is huge and is not required.\n\nTherefore\n\n I_2024 = 1012 (H_{2024} – 1).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.722020",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher–order structure:  The problem involves the second-order reciprocal sum  S_N(y), which is tied to the second derivative of a degree-2024 polynomial.  This is a full step beyond the first-order sum in both the original problem and the current kernel variant.  \n\n•  Deeper calculus–combinatorics link:  Solving the integral demands recognising and proving that S_N(y) equals ½ · P_N''(y)/P_N(y); this requires facility with elementary symmetric functions and higher derivatives, not merely a first derivative trick.  \n\n•  General-N derivation:  One must carry out the computation symbolically for arbitrary N before inserting N = 2024, adding an additional layer of abstraction.  \n\n•  Harmonic-number simplification:  The final evaluation hinges on manipulating harmonic numbers and telescoping identities, concepts that did not appear in the original versions.\n\n•  Outcome:  The calculation now spans polynomial algebra, higher-order derivatives, definite integration, and harmonic-number identities—multiple interacting ideas that collectively make the variant substantially more challenging than the original problem or its current kernel form."
      }
    },
    "original_kernel_variant": {
      "question": "Let N be a positive integer (later we will specialise to N = 2024).  \nFor α∈ℝ let  \n  C_N(α) be the coefficient of x^N in the Taylor expansion of (1+x)^α about x=0, i.e.  \n\n  C_N(α)=α(α−1)⋯(α−N+1)/N!.\n\nFor y>−1 put  \n\n  S_N(y)=∑_{1≤i<j≤N} 1/[(y+i)(y+j)].\n\nEvaluate the definite integral  \n\n  I_N = ∫_0^1 C_N(−y−1)·S_N(y) dy\n\nin closed form for general N, and then give the explicit value when N = 2024.",
      "solution": "Step 1.  Write the coefficient as a factored polynomial.  \nFor all integers N≥1,\n\n C_N(−y−1)= (−1)^N (y+1)(y+2)⋯(y+N)/N!\n      = (−1)^N P_N(y)/N!,  where P_N(y)=∏_{k=1}^N (y+k).\n\nStep 2.  Express S_N(y) through derivatives of P_N.  \nBecause P_N′(y)=P_N(y)∑_{k=1}^N 1/(y+k), we have  \n\n P_N″(y)=P_N(y) [(∑_{k=1}^N 1/(y+k))^2 − ∑_{k=1}^N 1/(y+k)^2]\n     = 2 P_N(y) S_N(y).\n\nThus\n\n S_N(y)=P_N″(y)/(2P_N(y)).                     (1)\n\nStep 3.  Insert (1) into the integral.  \nUsing C_N(−y−1)=(−1)^N P_N(y)/N!,\n\n C_N(−y−1)·S_N(y)= (−1)^N P_N″(y)/(2N!).\n\nHence  \n\n I_N = ∫_0^1 (−1)^N P_N″(y)/(2N!) dy\n    = (−1)^N [P_N′(y)]_0^1 /(2N!).    (2)\n\nStep 4.  Evaluate P_N′ at y=0 and y=1.  \nWe again use P_N′(y)=P_N(y)∑_{k=1}^N 1/(y+k).\n\n•  y=0: P_N(0)=N! ⇒ P_N′(0)=N!·H_N,  \n where H_N=∑_{k=1}^N 1/k is the N-th harmonic number.\n\n•  y=1: P_N(1)=(N+1)! and ∑_{k=1}^N 1/(1+k)=H_{N+1}−1,  \n so P_N′(1)=(N+1)! (H_{N+1}−1).\n\nInsert these into (2):\n\n I_N = (−1)^N[(N+1)! (H_{N+1}−1)−N! H_N]/(2N!)\n    = (−1)^N[(N+1)(H_{N+1}−1)−H_N]/2.     (3)\n\nStep 5.  Remove H_{N+1}.  \nSince H_{N+1}=H_N+1/(N+1),\n\n (N+1)(H_{N+1}−1)−H_N\n = (N+1)(H_N−1)+1−H_N\n = N H_N−N.\n\nSubstituting in (3) yields the compact formula\n\n I_N = (−1)^N · N/2 · (H_N − 1).              (4)\n\nStep 6.  Specialise to N = 2024.  \nBecause 2024 is even, (−1)^2024 = +1, so\n\n I_2024 = 1012·(H_{2024} − 1),\n\nwhere H_{2024}=∑_{k=1}^{2024} 1/k.  This is a rational number with denominator lcm{1,2,…,2024}; its explicit fractional form is huge and is not required.\n\nTherefore\n\n I_2024 = 1012 (H_{2024} – 1).",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.561580",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher–order structure:  The problem involves the second-order reciprocal sum  S_N(y), which is tied to the second derivative of a degree-2024 polynomial.  This is a full step beyond the first-order sum in both the original problem and the current kernel variant.  \n\n•  Deeper calculus–combinatorics link:  Solving the integral demands recognising and proving that S_N(y) equals ½ · P_N''(y)/P_N(y); this requires facility with elementary symmetric functions and higher derivatives, not merely a first derivative trick.  \n\n•  General-N derivation:  One must carry out the computation symbolically for arbitrary N before inserting N = 2024, adding an additional layer of abstraction.  \n\n•  Harmonic-number simplification:  The final evaluation hinges on manipulating harmonic numbers and telescoping identities, concepts that did not appear in the original versions.\n\n•  Outcome:  The calculation now spans polynomial algebra, higher-order derivatives, definite integration, and harmonic-number identities—multiple interacting ideas that collectively make the variant substantially more challenging than the original problem or its current kernel form."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}