CAIE FP1 2011 November — Question 1 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeInfinite series convergence and sum
DifficultyChallenging +1.2 This is a telescoping series question requiring verification of an algebraic identity, summation using the telescoping property, and finding a limit. While it involves multiple steps and the concept of series convergence, the techniques are standard for Further Maths: the telescoping pattern is given explicitly, the summation collapses straightforwardly, and the final part is routine inequality manipulation. It's harder than average A-level due to being Further Maths content, but represents a typical textbook exercise in this topic rather than requiring novel insight.
Spec4.06b Method of differences: telescoping series8.01a Recurrence relations: general sequences, closed form and recurrence8.01b Induction: prove results for sequences and series8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states
1 Verify that \(\frac { 1 } { n ^ { 2 } } - \frac { 1 } { ( n + 1 ) ^ { 2 } } = \frac { 2 n + 1 } { n ^ { 2 } ( n + 1 ) ^ { 2 } }\). Let \(S _ { N } = \sum _ { r = 1 } ^ { N } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }\). Express \(S _ { N }\) in terms of \(N\). Let \(S = \lim _ { N \rightarrow \infty } S _ { N }\). Find the least value of \(N\) such that \(S - S _ { N } < 10 ^ { - 16 }\).
Question 1:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{n^2+2n+1-n^2}{n^2(n+1)^2} = \frac{2n+1}{n^2(n+1)^2}\)B1 Verifies result (AG)
\(S_N = \left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\cdots+\left(\frac{1}{N^2}-\frac{1}{(N+1)^2}\right)\)M1 Uses difference method
\(= 1 - \frac{1}{(N+1)^2}\)A1 To sum; Part marks: 2
\(S - S_N < 10^{-16} \Rightarrow \frac{1}{(N+1)^2} < 10^{-16}\)M1 Considers difference between sum and sum to infinity
\(\Rightarrow (N+1) > 10^8\)A1
\(\Rightarrow\) least \(N = 10^8\)A1 Part marks: 3; Total: [6] 
## Question 1:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{n^2} - \frac{1}{(n+1)^2} = \frac{n^2+2n+1-n^2}{n^2(n+1)^2} = \frac{2n+1}{n^2(n+1)^2}$ | B1 | Verifies result (AG) |
| $S_N = \left(\frac{1}{1^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^2}-\frac{1}{3^2}\right)+\cdots+\left(\frac{1}{N^2}-\frac{1}{(N+1)^2}\right)$ | M1 | Uses difference method |
| $= 1 - \frac{1}{(N+1)^2}$ | A1 | To sum; Part marks: 2 |
| $S - S_N < 10^{-16} \Rightarrow \frac{1}{(N+1)^2} < 10^{-16}$ | M1 | Considers difference between sum and sum to infinity |
| $\Rightarrow (N+1) > 10^8$ | A1 | |
| $\Rightarrow$ least $N = 10^8$ | A1 | Part marks: 3; **Total: [6]** |

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1 Verify that $\frac { 1 } { n ^ { 2 } } - \frac { 1 } { ( n + 1 ) ^ { 2 } } = \frac { 2 n + 1 } { n ^ { 2 } ( n + 1 ) ^ { 2 } }$.

Let $S _ { N } = \sum _ { r = 1 } ^ { N } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }$. Express $S _ { N }$ in terms of $N$.

Let $S = \lim _ { N \rightarrow \infty } S _ { N }$. Find the least value of $N$ such that $S - S _ { N } < 10 ^ { - 16 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q1 [6]}}
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