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1989 Paper 3 Q10
D: 1700.0 B: 1516.0
summation proof by induction telescoping factorial products Riemann sum integration as area inequality definite integral from first principles

  1. Prove that \[ \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5) \] and deduce that \[ \sum_{r=1}^{n}r^{5}<\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5). \]
  2. Prove that, if \(n>1,\) \[ \sum_{r=0}^{n-1}r^{5}>\tfrac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n. \]
  3. Let \(\mathrm{f}\) be an increasing function. If the limits \[ \lim_{n\rightarrow\infty}\sum_{r=0}^{n-1}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right)\qquad\mbox{ and }\qquad\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\frac{a}{n}\mathrm{f}\left(\frac{ra}{n}\right) \] both exist and are equal, the definite integral \({\displaystyle \int_{0}^{a}\mathrm{f}(x)\,\mathrm{d}x}\) is defined to be their common value. Using this definition, prove that \[ \int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6. \]

Solution:

  1. Claim: \[ \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)=\tfrac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5) \] Proof: (By Induction) Base case: (n=1) \begin{align*} LHS &= 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 5! \\ RHS &= \frac16 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 5! \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n=k\), then consider \(n = k+1\) \begin{align*} \sum_{r=1}^{k+1} r(r+1)(r+2)(r+3)(r+4) &= \sum_{r=1}^{k} r(r+1)(r+2)(r+3)(r+4) + (k+1)(k+2)(k+3)(k+4)(k+5) \\ &\underbrace{=}_{\text{assumption}} \frac16 k(k+1)(k+2)(k+3)(k+4)(k+5) + (k+1)(k+2)(k+3)(k+4)(k+5) \\ &= (k+1)(k+2)(k+3)(k+4)(k+5) \l \frac{k}{6} +1\r \\ &= \frac16 (k+1)(k+2)(k+3)(k+4)(k+5)(k+6) \end{align*} Therefore our statement is true for \(n = k+1\). Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) then it is true for \(n = k+1\) by the principle of mathematical induction it is true for all \(n \geq 1\) Since \begin{align*} \sum_{r=1}^{n}r^5 &< \sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4) \\ &= \frac16 n(n+1)(n+2)(n+3)(n+4)(n+5) \end{align*}
  2. \begin{align*}\sum_{r=0}^{n-1} r^5 &> \sum_{r=0}^{n-1} (r-4)(r-3)(r-2)(r-1)r \\ &= \sum_{r=0}^{n-5} r(r+1)(r+2)(r+3)(r+4) \\ &= \frac16 (n-5)(n-4)(n-3)(n-2)(n-1)n \end{align*}
  3. Let \(f(x) = x^5\) \begin{align*} S_{1,n} &= \sum_{r=0}^{n-1}\frac{a}{n}f\left(\frac{ra}{n}\right) \\ &= \sum_{r=0}^{n-1}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\ &=\frac{a^6}{n^6} \sum_{r=0}^{n-1}r^5\\ \end{align*} Therefore \(\frac{a^6}6 \frac{(n-5)(n-4)(n-3)(n-2)(n-1)n}{n^6} < S_{1,n} < \frac{a^6}6 \frac{(n-1)n(n+1)(n+2)(n+3)(n+4)}{n^6}\) and so \(\lim_{n\to\infty} S_{1,n} = \frac{a^6}{6}\). Similarly, \begin{align*} S_{2,n} &= \sum_{r=1}^{n}\frac{a}{n}f\left(\frac{ra}{n}\right) \\ &= \sum_{r=1}^{n}\frac{a}{n}\left(\frac{ra}{n}\right)^5 \\ &= \frac{a^6}{n^6} \sum_{r=1}^{n} r^5 \end{align*} Therefore \(\frac{a^6}6 \frac{(n-4)(n-3)(n-2)(n-1)n(n+1)}{n^6} < S_{2,n} < \frac{a^6}6 \frac{n(n+1)(n+2)(n+3)(n+4)(n+5)}{n^6}\) and so \(\lim_{n\to\infty} S_{2,n} = \frac{a^6}{6}\). Since both limits exist are are equal, we have \[ \int_{0}^{a}x^{5}\,\mathrm{d}x=\tfrac{1}{6}a^6. \]

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