{
  "index": "1958-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. What is the smallest amount that may be invested at interest rate \\( i \\), compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, \\( \\ldots, n^{2} \\) dollars at the end of the \\( n \\)th year, in perpetuity?",
  "solution": "Solution. The present value of one dollar to be paid after \\( n \\) years is \\( (1+i)^{-n} \\) dollars. Hence the value in dollars of the given annuity is\n\\[\n\\sum_{n=1}^{\\infty} n^{2}(1+i)^{-n}\n\\]\n\nSince\n\\[\n\\frac{1}{1-x}=\\sum_{n}^{\\infty} x^{n}\n\\]\nwe have\n\\[\n\\frac{x}{(1-x)^{2}}=x \\frac{d}{d x}\\left(\\frac{1}{1-x}\\right)=\\sum_{n=1}^{\\infty} n x^{\\prime \\prime}\n\\]\nand\n\\[\n\\frac{x+x^{2}}{(1-x)^{3}}=x \\frac{d}{d x}\\left(\\frac{x}{(1-x)^{2}}\\right)=\\sum_{n=1}^{\\infty} n^{2} x^{n}\n\\]\nfor \\( |x|<1 \\). Putting \\( x=1 /(1+i) \\), we obtain\n\\[\n\\sum_{n=1}^{\\infty} n^{2}(1+i)^{-n}=\\frac{(1+i)(2+i)}{i^{3}}\n\\]\n(At 6 percent interest, the cost of the annuity would be \\( \\$ 10,109.26 \\).)",
  "vars": [
    "n",
    "x"
  ],
  "params": [
    "i"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "yearindex",
        "x": "seriesvar",
        "i": "interestrate"
      },
      "question": "6. What is the smallest amount that may be invested at interest rate \\( interestrate \\), compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, \\( \\ldots, yearindex^{2} \\) dollars at the end of the \\( yearindex \\)th year, in perpetuity?",
      "solution": "Solution. The present value of one dollar to be paid after \\( yearindex \\) years is \\( (1+interestrate)^{-yearindex} \\) dollars. Hence the value in dollars of the given annuity is\n\\[\n\\sum_{yearindex=1}^{\\infty} yearindex^{2}(1+interestrate)^{-yearindex}\n\\]\n\nSince\n\\[\n\\frac{1}{1-seriesvar}=\\sum_{yearindex}^{\\infty} seriesvar^{yearindex}\n\\]\nwe have\n\\[\n\\frac{seriesvar}{(1-seriesvar)^{2}}=seriesvar \\frac{d}{d seriesvar}\\left(\\frac{1}{1-seriesvar}\\right)=\\sum_{yearindex=1}^{\\infty} yearindex seriesvar^{\\prime \\prime}\n\\]\nand\n\\[\n\\frac{seriesvar+seriesvar^{2}}{(1-seriesvar)^{3}}=seriesvar \\frac{d}{d seriesvar}\\left(\\frac{seriesvar}{(1-seriesvar)^{2}}\\right)=\\sum_{yearindex=1}^{\\infty} yearindex^{2} seriesvar^{yearindex}\n\\]\nfor \\( |seriesvar|<1 \\). Putting \\( seriesvar=1 /(1+interestrate) \\), we obtain\n\\[\n\\sum_{yearindex=1}^{\\infty} yearindex^{2}(1+interestrate)^{-yearindex}=\\frac{(1+interestrate)(2+interestrate)}{interestrate^{3}}\n\\]\n(At 6 percent interest, the cost of the annuity would be \\( \\$ 10,109.26 \\).)"
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "pineapple",
        "x": "marigold"
      },
      "question": "6. What is the smallest amount that may be invested at interest rate \\( i \\), compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, \\( \\ldots, pineapple^{2} \\) dollars at the end of the \\( pineapple \\)th year, in perpetuity?",
      "solution": "Solution. The present value of one dollar to be paid after \\( pineapple \\) years is \\( (1+i)^{-pineapple} \\) dollars. Hence the value in dollars of the given annuity is\n\\[\n\\sum_{pineapple=1}^{\\infty} pineapple^{2}(1+i)^{-pineapple}\n\\]\n\nSince\n\\[\n\\frac{1}{1-marigold}=\\sum_{pineapple}^{\\infty} marigold^{pineapple}\n\\]\nwe have\n\\[\n\\frac{marigold}{(1-marigold)^{2}}=marigold \\frac{d}{d marigold}\\left(\\frac{1}{1-marigold}\\right)=\\sum_{pineapple=1}^{\\infty} pineapple marigold^{\\prime \\prime}\n\\]\nand\n\\[\n\\frac{marigold+marigold^{2}}{(1-marigold)^{3}}=marigold \\frac{d}{d marigold}\\left(\\frac{marigold}{(1-marigold)^{2}}\\right)=\\sum_{pineapple=1}^{\\infty} pineapple^{2} marigold^{pineapple}\n\\]\nfor \\( |marigold|<1 \\). Putting \\( marigold=1 /(1+i) \\), we obtain\n\\[\n\\sum_{pineapple=1}^{\\infty} pineapple^{2}(1+i)^{-pineapple}=\\frac{(1+i)(2+i)}{i^{3}}\n\\]\n(At 6 percent interest, the cost of the annuity would be \\$ 10,109.26 .)"
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "noninteger",
        "x": "constantvalue",
        "i": "lossrate"
      },
      "question": "6. What is the smallest amount that may be invested at interest rate \\( lossrate \\), compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, \\( \\ldots, noninteger^{2} \\) dollars at the end of the \\( noninteger \\)th year, in perpetuity?",
      "solution": "Solution. The present value of one dollar to be paid after \\( noninteger \\) years is \\( (1+lossrate)^{-noninteger} \\) dollars. Hence the value in dollars of the given annuity is\n\\[\n\\sum_{noninteger=1}^{\\infty} noninteger^{2}(1+lossrate)^{-noninteger}\n\\]\n\nSince\n\\[\n\\frac{1}{1-constantvalue}=\\sum_{noninteger}^{\\infty} constantvalue^{noninteger}\n\\]\nwe have\n\\[\n\\frac{constantvalue}{(1-constantvalue)^{2}}=constantvalue \\frac{d}{d constantvalue}\\left(\\frac{1}{1-constantvalue}\\right)=\\sum_{noninteger=1}^{\\infty} noninteger constantvalue^{\\prime \\prime}\n\\]\nand\n\\[\n\\frac{constantvalue+constantvalue^{2}}{(1-constantvalue)^{3}}=constantvalue \\frac{d}{d constantvalue}\\left(\\frac{constantvalue}{(1-constantvalue)^{2}}\\right)=\\sum_{noninteger=1}^{\\infty} noninteger^{2} constantvalue^{noninteger}\n\\]\nfor \\( |constantvalue|<1 \\). Putting \\( constantvalue=1 /(1+lossrate) \\), we obtain\n\\[\n\\sum_{noninteger=1}^{\\infty} noninteger^{2}(1+lossrate)^{-noninteger}=\\frac{(1+lossrate)(2+lossrate)}{lossrate^{3}}\n\\]\n(At 6 percent interest, the cost of the annuity would be \\( \\$ 10,109.26 \\).)"
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "x": "hjgrksla"
      },
      "question": "6. What is the smallest amount that may be invested at interest rate \\( i \\), compounded annually, in order that one may withdraw 1 dollar at the end of the first year, 4 dollars at the end of the second year, \\( \\ldots, qzxwvtnp^{2} \\) dollars at the end of the \\( qzxwvtnp \\)th year, in perpetuity?",
      "solution": "Solution. The present value of one dollar to be paid after \\( qzxwvtnp \\) years is \\( (1+i)^{-qzxwvtnp} \\) dollars. Hence the value in dollars of the given annuity is\n\\[\n\\sum_{qzxwvtnp=1}^{\\infty} qzxwvtnp^{2}(1+i)^{-qzxwvtnp}\n\\]\n\nSince\n\\[\n\\frac{1}{1-hjgrksla}=\\sum_{qzxwvtnp}^{\\infty} hjgrksla^{qzxwvtnp}\n\\]\nwe have\n\\[\n\\frac{hjgrksla}{(1-hjgrksla)^{2}}=hjgrksla \\frac{d}{d hjgrksla}\\left(\\frac{1}{1-hjgrksla}\\right)=\\sum_{qzxwvtnp=1}^{\\infty} qzxwvtnp hjgrksla^{\\prime \\prime}\n\\]\nand\n\\[\n\\frac{hjgrksla+hjgrksla^{2}}{(1-hjgrksla)^{3}}=hjgrksla \\frac{d}{d hjgrksla}\\left(\\frac{hjgrksla}{(1-hjgrksla)^{2}}\\right)=\\sum_{qzxwvtnp=1}^{\\infty} qzxwvtnp^{2} hjgrksla^{qzxwvtnp}\n\\]\nfor \\( |hjgrksla|<1 \\). Putting \\( hjgrksla=1 /(1+i) \\), we obtain\n\\[\n\\sum_{qzxwvtnp=1}^{\\infty} qzxwvtnp^{2}(1+i)^{-qzxwvtnp}=\\frac{(1+i)(2+i)}{i^{3}}\n\\]\n(At 6 percent interest, the cost of the annuity would be \\( \\$ 10,109.26 \\).)"
    },
    "kernel_variant": {
      "question": "---------------------------------------------------------------------------\nA philanthropic foundation places a single deposit at time 0 into a fund that credits an effective annual return which repeats with a 3-year cycle\n\n Year 1 earns i₁ (the balance is multiplied by 1+i₁),  \n Year 2 earns i₂,  \n Year 3 earns i₃,\n\nand thereafter the pattern (i₁,i₂,i₃) repeats indefinitely (year 4 again earns i₁, etc.).  \nAssume i₁,i₂,i₃ are positive and satisfy\n\n (1+i₁)(1+i₂)(1+i₃)  >  (1+g)³ ,  g>0,\n\nso that the average geometric return exceeds the fixed annual inflation rate g.\n\nStarting one year from now the foundation must make an endless stream of nominal withdrawals\n\n Wₙ = Tₙ·(1+g)ⁿ  (n = 1,2,3,…),\n\nwhere Tₙ = n(n+1)/2 is the n-th triangular number (T₁=1, T₂=3, T₃=6,…).  \nEach withdrawal occurs at the end of the year, immediately after that year’s interest has been credited.\n\nFind, in closed form, the minimum initial deposit that will finance all withdrawals forever, expressed in terms of i₁,i₂,i₃ and g.\n\n---------------------------------------------------------------------------",
      "solution": "---------------------------------------------------------------------------\n1.  Discount notation  \n   d₁ = (1+i₁)⁻¹, d₂ = (1+i₂)⁻¹, d₃ = (1+i₃)⁻¹  (one-year discount factors)  \n   q  = 1+g             (inflation multiplier, q>1)\n\n2.  Separate the cash flow into the three residue classes mod 3  \n   Write n = 3k+r with k ∈ ℕ₀ and r ∈ {1,2,3}.  \n   The nominal withdrawal in year n is Wₙ = Tₙ qⁿ.\n\n   Discount factor for year n  \n   The accumulation over the first n=3k+r years is\n\n  (1+i₁)^{\\,k+δ₁(r)}(1+i₂)^{\\,k+δ₂(r)}(1+i₃)^{\\,k+δ₃(r)}\n\n   where  \n  δ₁(r)=1 if r≥1 else 0, δ₂(r)=1 if r≥2 else 0, δ₃(r)=1 if r=3 else 0.\n\n   Hence the present-value multiplier for year n is  \n\n  Dₙ = d₁^{\\,k+δ₁(r)} d₂^{\\,k+δ₂(r)} d₃^{\\,k+δ₃(r)}.\n\n   Introduce the three-year factors  \n\n  A := d₁d₂d₃  (one complete cycle discount),  \n  θ_r := d₁^{δ₁(r)}d₂^{δ₂(r)}d₃^{δ₃(r)}  (r=1,2,3).\n\n   Then D_{3k+r}=θ_r A^{k}.  Note θ₁=d₁, θ₂=d₁d₂, θ₃=A.\n\n3.  Present value (PV) of the withdrawal stream  \n   PV = Σ_{n≥1} Tₙ qⁿ Dₙ  \n   = Σ_{k≥0} Σ_{r=1}^{3} T_{3k+r} q^{\\,3k+r} θ_r A^{k}.\n\n   Put B := q³A.  Because (1+i₁)(1+i₂)(1+i₃) > q³, we have B<1, so all series converge.\n\n   Define\n\n  S_r := Σ_{k=0}^{∞} T_{3k+r} B^{k},  r=1,2,3. (1)\n\n   Then PV = θ₁ q S₁ + θ₂ q² S₂ + θ₃ q³ S₃.\n\n4.  Expansion of the triangular numbers  \n   For fixed r write T_{3k+r} explicitly:\n\n  T_{3k+r} = (3k+r)(3k+r+1)/2  \n       = (9/2)k² + 3k(r+½) + r(r+1)/2.    (2)\n\n   Therefore\n\n  S_r = Σ_{k≥0} (α k² + β_r k + γ_r) B^{k},\n\n   with unified constants  \n\n  α = 9/2,  \n  β_r = 3(r+½) = 3r + 3/2,  \n  γ_r = r(r+1)/2.                         (3)\n\n5.  Geometric–power sums (|B|<1)\n\n Σ_{k≥0} B^{k} = 1/(1−B),  \n Σ_{k≥0} k B^{k} = B/(1−B)²,  \n Σ_{k≥0} k² B^{k} = B(1+B)/(1−B)³.\n\n   Substitute (3):\n\n  S_r = α· B(1+B)/(1−B)³ + β_r· B/(1−B)² + γ_r· 1/(1−B).    (4)\n\n6.  Evaluate β_r, γ_r once and for all\n\n r = 1: β₁ = 9/2, γ₁ = 1  \n r = 2: β₂ = 15/2, γ₂ = 3  \n r = 3: β₃ = 21/2, γ₃ = 6.\n\n   Because α = 9/2 for all r, (4) becomes\n\n  S_r = (9/2)· B(1+B)/(1−B)³ + β_r· B/(1−B)² + γ_r/(1−B).   (5)\n\n7.  Assemble PV  \n   PV = θ₁ q S₁ + θ₂ q² S₂ + θ₃ q³ S₃, with θ₁=d₁, θ₂=d₁d₂, θ₃=A.\n\n   Introduce the common kernels\n\n  K₀ := 1/(1−B), K₁ := B/(1−B)², K₂ := B(1+B)/(1−B)³.\n\n   Factoring (5):\n\n PV = (9/2) K₂ (q d₁ + q² d₁d₂ + q³ A)  \n    + K₁ ( (9/2)q d₁ + (15/2)q² d₁d₂ + (21/2)q³ A )  \n    + K₀ (       q d₁ +       3 q² d₁d₂ +        6 q³ A ).   (6)\n\n8.  Replace the discounts with the interest rates  \n   d₁ = (1+i₁)⁻¹, d₂ = (1+i₂)⁻¹, d₃ = (1+i₃)⁻¹,  \n   A   = d₁d₂d₃,  B   = q³A with q = 1+g.\n\n   Expression (6) is therefore a rational function of i₁,i₂,i₃ and g and equals the minimum initial deposit required.\n\n   Any equivalent algebraic simplification of (6) is acceptable as the final closed form.\n\n---------------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.506734",
        "was_fixed": false,
        "difficulty_analysis": "---------------------------------------------------------------------------  \n• Variable Discount Structure:  The interest rate now follows a 3-year cycle (i₁,i₂,i₃).  The discount factors are not a simple geometric sequence; they must be grouped by congruence classes, producing three intertwined series instead of one.\n\n• Inflation Adjustment:  Each withdrawal is multiplied by (1+g)ⁿ, introducing an additional geometric growth that interacts non-trivially with the discount cycle.\n\n• Polynomial Cash-Flows:  The payments grow according to the triangular numbers Tₙ = n(n+1)/2, a quadratic sequence that forces the evaluator to handle sums of k , k² terms rather than the linear or quadratic cases of the original problem.\n\n• Multi-Layered Series Decomposition:  Solving the problem requires separating the infinite series into three coupled sub-series, computing power-series sums of quadratics, and then recombining them—a clear increase in algebraic and organisational complexity.\n\n• Closed-Form Requirement:  Producing a single rational expression in i₁,i₂,i₃,g demands careful symbolic manipulation far beyond the straight derivative-of-geometric-series trick that settles the original n²-annuity.\n\nAltogether, the enhanced variant requires deeper insight into periodic discounting, inflation, and higher-degree generating functions, thus raising both conceptual and computational difficulty well above the original."
      }
    },
    "original_kernel_variant": {
      "question": "---------------------------------------------------------------------------\nA philanthropic foundation places a single deposit at time 0 into a fund that credits an effective annual return which repeats with a 3-year cycle\n\n Year 1 earns i₁ (the balance is multiplied by 1+i₁),  \n Year 2 earns i₂,  \n Year 3 earns i₃,\n\nand thereafter the pattern (i₁,i₂,i₃) repeats indefinitely (year 4 again earns i₁, etc.).  \nAssume i₁,i₂,i₃ are positive and satisfy\n\n (1+i₁)(1+i₂)(1+i₃)  >  (1+g)³ ,  g>0,\n\nso that the average geometric return exceeds the fixed annual inflation rate g.\n\nStarting one year from now the foundation must make an endless stream of nominal withdrawals\n\n Wₙ = Tₙ·(1+g)ⁿ  (n = 1,2,3,…),\n\nwhere Tₙ = n(n+1)/2 is the n-th triangular number (T₁=1, T₂=3, T₃=6,…).  \nEach withdrawal occurs at the end of the year, immediately after that year’s interest has been credited.\n\nFind, in closed form, the minimum initial deposit that will finance all withdrawals forever, expressed in terms of i₁,i₂,i₃ and g.\n\n---------------------------------------------------------------------------",
      "solution": "---------------------------------------------------------------------------\n1.  Discount notation  \n   d₁ = (1+i₁)⁻¹, d₂ = (1+i₂)⁻¹, d₃ = (1+i₃)⁻¹  (one-year discount factors)  \n   q  = 1+g             (inflation multiplier, q>1)\n\n2.  Separate the cash flow into the three residue classes mod 3  \n   Write n = 3k+r with k ∈ ℕ₀ and r ∈ {1,2,3}.  \n   The nominal withdrawal in year n is Wₙ = Tₙ qⁿ.\n\n   Discount factor for year n  \n   The accumulation over the first n=3k+r years is\n\n  (1+i₁)^{\\,k+δ₁(r)}(1+i₂)^{\\,k+δ₂(r)}(1+i₃)^{\\,k+δ₃(r)}\n\n   where  \n  δ₁(r)=1 if r≥1 else 0, δ₂(r)=1 if r≥2 else 0, δ₃(r)=1 if r=3 else 0.\n\n   Hence the present-value multiplier for year n is  \n\n  Dₙ = d₁^{\\,k+δ₁(r)} d₂^{\\,k+δ₂(r)} d₃^{\\,k+δ₃(r)}.\n\n   Introduce the three-year factors  \n\n  A := d₁d₂d₃  (one complete cycle discount),  \n  θ_r := d₁^{δ₁(r)}d₂^{δ₂(r)}d₃^{δ₃(r)}  (r=1,2,3).\n\n   Then D_{3k+r}=θ_r A^{k}.  Note θ₁=d₁, θ₂=d₁d₂, θ₃=A.\n\n3.  Present value (PV) of the withdrawal stream  \n   PV = Σ_{n≥1} Tₙ qⁿ Dₙ  \n   = Σ_{k≥0} Σ_{r=1}^{3} T_{3k+r} q^{\\,3k+r} θ_r A^{k}.\n\n   Put B := q³A.  Because (1+i₁)(1+i₂)(1+i₃) > q³, we have B<1, so all series converge.\n\n   Define\n\n  S_r := Σ_{k=0}^{∞} T_{3k+r} B^{k},  r=1,2,3. (1)\n\n   Then PV = θ₁ q S₁ + θ₂ q² S₂ + θ₃ q³ S₃.\n\n4.  Expansion of the triangular numbers  \n   For fixed r write T_{3k+r} explicitly:\n\n  T_{3k+r} = (3k+r)(3k+r+1)/2  \n       = (9/2)k² + 3k(r+½) + r(r+1)/2.    (2)\n\n   Therefore\n\n  S_r = Σ_{k≥0} (α k² + β_r k + γ_r) B^{k},\n\n   with unified constants  \n\n  α = 9/2,  \n  β_r = 3(r+½) = 3r + 3/2,  \n  γ_r = r(r+1)/2.                         (3)\n\n5.  Geometric–power sums (|B|<1)\n\n Σ_{k≥0} B^{k} = 1/(1−B),  \n Σ_{k≥0} k B^{k} = B/(1−B)²,  \n Σ_{k≥0} k² B^{k} = B(1+B)/(1−B)³.\n\n   Substitute (3):\n\n  S_r = α· B(1+B)/(1−B)³ + β_r· B/(1−B)² + γ_r· 1/(1−B).    (4)\n\n6.  Evaluate β_r, γ_r once and for all\n\n r = 1: β₁ = 9/2, γ₁ = 1  \n r = 2: β₂ = 15/2, γ₂ = 3  \n r = 3: β₃ = 21/2, γ₃ = 6.\n\n   Because α = 9/2 for all r, (4) becomes\n\n  S_r = (9/2)· B(1+B)/(1−B)³ + β_r· B/(1−B)² + γ_r/(1−B).   (5)\n\n7.  Assemble PV  \n   PV = θ₁ q S₁ + θ₂ q² S₂ + θ₃ q³ S₃, with θ₁=d₁, θ₂=d₁d₂, θ₃=A.\n\n   Introduce the common kernels\n\n  K₀ := 1/(1−B), K₁ := B/(1−B)², K₂ := B(1+B)/(1−B)³.\n\n   Factoring (5):\n\n PV = (9/2) K₂ (q d₁ + q² d₁d₂ + q³ A)  \n    + K₁ ( (9/2)q d₁ + (15/2)q² d₁d₂ + (21/2)q³ A )  \n    + K₀ (       q d₁ +       3 q² d₁d₂ +        6 q³ A ).   (6)\n\n8.  Replace the discounts with the interest rates  \n   d₁ = (1+i₁)⁻¹, d₂ = (1+i₂)⁻¹, d₃ = (1+i₃)⁻¹,  \n   A   = d₁d₂d₃,  B   = q³A with q = 1+g.\n\n   Expression (6) is therefore a rational function of i₁,i₂,i₃ and g and equals the minimum initial deposit required.\n\n   Any equivalent algebraic simplification of (6) is acceptable as the final closed form.\n\n---------------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.423252",
        "was_fixed": false,
        "difficulty_analysis": "---------------------------------------------------------------------------  \n• Variable Discount Structure:  The interest rate now follows a 3-year cycle (i₁,i₂,i₃).  The discount factors are not a simple geometric sequence; they must be grouped by congruence classes, producing three intertwined series instead of one.\n\n• Inflation Adjustment:  Each withdrawal is multiplied by (1+g)ⁿ, introducing an additional geometric growth that interacts non-trivially with the discount cycle.\n\n• Polynomial Cash-Flows:  The payments grow according to the triangular numbers Tₙ = n(n+1)/2, a quadratic sequence that forces the evaluator to handle sums of k , k² terms rather than the linear or quadratic cases of the original problem.\n\n• Multi-Layered Series Decomposition:  Solving the problem requires separating the infinite series into three coupled sub-series, computing power-series sums of quadratics, and then recombining them—a clear increase in algebraic and organisational complexity.\n\n• Closed-Form Requirement:  Producing a single rational expression in i₁,i₂,i₃,g demands careful symbolic manipulation far beyond the straight derivative-of-geometric-series trick that settles the original n²-annuity.\n\nAltogether, the enhanced variant requires deeper insight into periodic discounting, inflation, and higher-degree generating functions, thus raising both conceptual and computational difficulty well above the original."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}