Pre-U Pre-U 9795/1 2017 June — Question 10 10 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2017
SessionJune
Marks10
TopicSequences and series, recurrence and convergence
TypeIntegral bounds for series
DifficultyChallenging +1.8 This is a sophisticated multi-part question requiring partial fractions, telescoping series, inequality reasoning, and integral bounds for series convergence. While each individual technique is A-level standard, part (iv) requires creative insight to bound an infinite series using the telescoping result from part (ii), which elevates it significantly above routine exercises. The combination of proof elements and non-standard application makes this harder than typical A-level questions but not exceptionally difficult for Further Maths students.
Spec1.02y Partial fractions: decompose rational functions4.06b Method of differences: telescoping series
10
  1. Express \(\frac { 1 } { ( k - 1 ) k ( k + 1 ) }\) in partial fractions.
  2. Let \(S _ { n } = \sum _ { k = 3 } ^ { n } \frac { 1 } { ( k - 1 ) k ( k + 1 ) }\) for \(n \geqslant 3\). Use the method of differences to show that $$S _ { n } = \frac { 1 } { 12 } - \frac { 1 } { 2 n ( n + 1 ) }$$ and write down the limit of \(S _ { n }\) as \(n \rightarrow \infty\).
  3. Given that \(k\) is a positive integer greater than 1 , explain why \(\frac { 1 } { k ^ { 3 } } < \frac { 1 } { ( k - 1 ) k ( k + 1 ) }\).
  4. Show that \(\frac { 27 } { 24 } < \sum _ { k = 1 } ^ { \infty } \frac { 1 } { k ^ { 3 } } < \frac { 29 } { 24 }\).
Question 10(i)
\(\dfrac{1}{(k-1)k(k+1)} \equiv \dfrac{A}{k-1} + \dfrac{B}{k} + \dfrac{C}{k+1}\) M1 Correct form
Equating terms / substitution / cover-up M1 Method for determining constants
\(\equiv \dfrac{\frac{1}{2}}{k-1} - \dfrac{1}{k} + \dfrac{\frac{1}{2}}{k+1}\) A1
Question 10(ii)
\(\displaystyle\sum_{k=3}^{n}\dfrac{1}{(k-1)k(k+1)} \equiv \dfrac{1}{2}\sum_{k=3}^{n}\dfrac{1}{k-1} + \dfrac{1}{2}\sum_{k=3}^{n}\dfrac{1}{k+1} - \sum_{k=3}^{n}\dfrac{1}{k}\) M1 Splitting up
\(\equiv \frac{1}{2}\!\left\{\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}\right\} + \frac{1}{2}\!\left\{\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\right\} - \left\{\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}\right\}\) M1 Attempt at cancelling of terms
\(\equiv \frac{1}{2}\!\left\{\frac{1}{2}+\frac{1}{3}\right\} + \frac{1}{2}\!\left\{\frac{1}{n}+\frac{1}{n+1}\right\} - \left\{\frac{1}{3}+\frac{1}{n}\right\}\) A1 Correct ones clearly identified
\(\equiv \dfrac{1}{12} - \dfrac{1}{2}\!\left\{\dfrac{1}{n} - \dfrac{1}{n+1}\right\} \equiv \dfrac{1}{12} - \dfrac{1}{2n(n+1)}\) A1 Legitimately shown (AG)
Limit \((S_n)\) as \(n\to\infty\) is \(S = \dfrac{1}{12}\) B1 FT
Question 10(iii)
\(k^3 > k^3 - k = k(k-1)(k+1)\)
\(\Rightarrow \dfrac{1}{k^3} < \dfrac{1}{(k-1)k(k+1)}\) B1
Question 10(iv)
\(\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^3} > 1 + \dfrac{1}{8} = \dfrac{9}{8} = \dfrac{22}{24}\) B1 Given result justified
\(\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^3} = 1 + \dfrac{1}{8} + \displaystyle\sum_{k=3}^{\infty}\dfrac{1}{k^3} < 1 + \dfrac{1}{8} + \displaystyle\sum_{k=3}^{n}\dfrac{1}{(k-1)k(k+1)}\) M1
\(= 1 + \dfrac{1}{8} + \dfrac{1}{12} = \dfrac{29}{24}\) A1 Given result justified
Total: 10 marks
**Question 10(i)**

$\dfrac{1}{(k-1)k(k+1)} \equiv \dfrac{A}{k-1} + \dfrac{B}{k} + \dfrac{C}{k+1}$ **M1** Correct form

Equating terms / substitution / cover-up **M1** Method for determining constants

$\equiv \dfrac{\frac{1}{2}}{k-1} - \dfrac{1}{k} + \dfrac{\frac{1}{2}}{k+1}$ **A1**

**Question 10(ii)**

$\displaystyle\sum_{k=3}^{n}\dfrac{1}{(k-1)k(k+1)} \equiv \dfrac{1}{2}\sum_{k=3}^{n}\dfrac{1}{k-1} + \dfrac{1}{2}\sum_{k=3}^{n}\dfrac{1}{k+1} - \sum_{k=3}^{n}\dfrac{1}{k}$ **M1** Splitting up

$\equiv \frac{1}{2}\!\left\{\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}\right\} + \frac{1}{2}\!\left\{\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\right\} - \left\{\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n-1}+\frac{1}{n}\right\}$ **M1** Attempt at cancelling of terms

$\equiv \frac{1}{2}\!\left\{\frac{1}{2}+\frac{1}{3}\right\} + \frac{1}{2}\!\left\{\frac{1}{n}+\frac{1}{n+1}\right\} - \left\{\frac{1}{3}+\frac{1}{n}\right\}$ **A1** Correct ones clearly identified

$\equiv \dfrac{1}{12} - \dfrac{1}{2}\!\left\{\dfrac{1}{n} - \dfrac{1}{n+1}\right\} \equiv \dfrac{1}{12} - \dfrac{1}{2n(n+1)}$ **A1** Legitimately shown (AG)

Limit $(S_n)$ as $n\to\infty$ is $S = \dfrac{1}{12}$ **B1** FT

**Question 10(iii)**

$k^3 > k^3 - k = k(k-1)(k+1)$

$\Rightarrow \dfrac{1}{k^3} < \dfrac{1}{(k-1)k(k+1)}$ **B1**

**Question 10(iv)**

$\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^3} > 1 + \dfrac{1}{8} = \dfrac{9}{8} = \dfrac{22}{24}$ **B1** Given result justified

$\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^3} = 1 + \dfrac{1}{8} + \displaystyle\sum_{k=3}^{\infty}\dfrac{1}{k^3} < 1 + \dfrac{1}{8} + \displaystyle\sum_{k=3}^{n}\dfrac{1}{(k-1)k(k+1)}$ **M1**

$= 1 + \dfrac{1}{8} + \dfrac{1}{12} = \dfrac{29}{24}$ **A1** Given result justified

**Total: 10 marks**
10 (i) Express $\frac { 1 } { ( k - 1 ) k ( k + 1 ) }$ in partial fractions.\\
(ii) Let $S _ { n } = \sum _ { k = 3 } ^ { n } \frac { 1 } { ( k - 1 ) k ( k + 1 ) }$ for $n \geqslant 3$. Use the method of differences to show that

$$S _ { n } = \frac { 1 } { 12 } - \frac { 1 } { 2 n ( n + 1 ) }$$

and write down the limit of $S _ { n }$ as $n \rightarrow \infty$.\\
(iii) Given that $k$ is a positive integer greater than 1 , explain why $\frac { 1 } { k ^ { 3 } } < \frac { 1 } { ( k - 1 ) k ( k + 1 ) }$.\\
(iv) Show that $\frac { 27 } { 24 } < \sum _ { k = 1 } ^ { \infty } \frac { 1 } { k ^ { 3 } } < \frac { 29 } { 24 }$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2017 Q10 [10]}}
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