8. (a) Prove by induction that, for any positive integer \(n\),
$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$
(b) Using the formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\), show that
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + 3 r + 2 \right) = \frac { 1 } { 4 } n ( n + 2 ) \left( n ^ { 2 } + 7 \right)$$
(c) Hence evaluate \(\sum _ { r = 15 } ^ { 25 } \left( r ^ { 3 } + 3 r + 2 \right)\)
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Question 8:
Part (a):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\sum_{r=1}^{1} r^3 = 1^3 = 1\) and \(\frac{1}{4} \times 1^2 \times 2^2 = 1\) B1
Assume true for \(n=k\): \(\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2 + (k+1)^3\) B1
\(\frac{1}{4}(k+1)^2\left[k^2 + 4(k+1)\right] = \frac{1}{4}(k+1)^2(k+2)^2\) M1 A1
Correct method to identify \((k+1)^2\) as factor for M1; \(\frac{1}{4}(k+1)^2(k+2)^2\) for A1
True for \(n=k+1\) if true for \(n=k\). True for \(n=1\), \(\therefore\) true for all \(n\). A1cso
All three elements stated for final A1
Part (b):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\sum r^3 + 3\sum r + \sum 2 = \frac{1}{4}n^2(n+1)^2 + 3\left(\frac{1}{2}n(n+1)\right) + 2n\) B1, B1
\(= \frac{1}{4}n\left[n(n+1)^2 + 6(n+1) + 8\right]\) M1
Attempt to factorise by \(n\)
\(= \frac{1}{4}n\left[n^3 + 2n^2 + 7n + 14\right] = \frac{1}{4}n(n+2)(n^2+7)\) A1 A1cso
\(\frac{1}{4}\) and \(n^3+2n^2+7n+14\) for first A1
Part (c):
Answer Marks
Guidance
Answer/Working Marks
Guidance
\(\sum_{15}^{25} = \sum_{1}^{25} - \sum_{1}^{14}\) M1
With attempt to substitute into part (b)
\(= \frac{1}{4}(25\times27\times632) - \frac{1}{4}(14\times16\times203) = 106650 - 11368 = 95282\) A1
No working \(0/2\)
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## Question 8:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{r=1}^{1} r^3 = 1^3 = 1$ and $\frac{1}{4} \times 1^2 \times 2^2 = 1$ | B1 | |
| Assume true for $n=k$: $\sum_{r=1}^{k+1} r^3 = \frac{1}{4}k^2(k+1)^2 + (k+1)^3$ | B1 | |
| $\frac{1}{4}(k+1)^2\left[k^2 + 4(k+1)\right] = \frac{1}{4}(k+1)^2(k+2)^2$ | M1 A1 | Correct method to identify $(k+1)^2$ as factor for M1; $\frac{1}{4}(k+1)^2(k+2)^2$ for A1 |
| True for $n=k+1$ if true for $n=k$. True for $n=1$, $\therefore$ true for all $n$. | A1cso | All three elements stated for final A1 |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum r^3 + 3\sum r + \sum 2 = \frac{1}{4}n^2(n+1)^2 + 3\left(\frac{1}{2}n(n+1)\right) + 2n$ | B1, B1 | |
| $= \frac{1}{4}n\left[n(n+1)^2 + 6(n+1) + 8\right]$ | M1 | Attempt to factorise by $n$ |
| $= \frac{1}{4}n\left[n^3 + 2n^2 + 7n + 14\right] = \frac{1}{4}n(n+2)(n^2+7)$ | A1 A1cso | $\frac{1}{4}$ and $n^3+2n^2+7n+14$ for first A1 |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum_{15}^{25} = \sum_{1}^{25} - \sum_{1}^{14}$ | M1 | With attempt to substitute into part (b) |
| $= \frac{1}{4}(25\times27\times632) - \frac{1}{4}(14\times16\times203) = 106650 - 11368 = 95282$ | A1 | No working $0/2$ |
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8. (a) Prove by induction that, for any positive integer $n$,
$$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$
(b) Using the formulae for $\sum _ { r = 1 } ^ { n } r$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, show that
$$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + 3 r + 2 \right) = \frac { 1 } { 4 } n ( n + 2 ) \left( n ^ { 2 } + 7 \right)$$
(c) Hence evaluate $\sum _ { r = 15 } ^ { 25 } \left( r ^ { 3 } + 3 r + 2 \right)$\\
\hfill \mbox{\textit{Edexcel FP1 2010 Q8 [12]}}