3 Prove by mathematical induction that, for all positive integers \(n , \sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }\).
State the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }\).
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Question 3:
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Working/Answer Marks
Notes
\(H_k: \sum_{r=1}^{k} \frac{1}{(2r)^2-1} = \frac{k}{2k+1}\) is true for some integer \(k\) B1
\(\frac{1}{2^2-1} = \frac{1}{3} = \frac{1}{2\times1+1} \Rightarrow H_1\) is true B1
\(\frac{k}{2k+1} + \frac{1}{(2k+2)^2-1} = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} = \frac{2k^2+3k+1}{(2k+1)(2k+3)}\) M1A1
\(= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2[k+1]+1}\) A1
\(\therefore H_k \Rightarrow H_{k+1}\); by PMI \(H_n\) true for all positive integers \(n\) A1
Requires all previous marks
\(\sum_{r=1}^{\infty} \frac{1}{(2r)^2-1} = \frac{1}{2}\) B1
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## Question 3:
| Working/Answer | Marks | Notes |
|---|---|---|
| $H_k: \sum_{r=1}^{k} \frac{1}{(2r)^2-1} = \frac{k}{2k+1}$ is true for some integer $k$ | B1 | |
| $\frac{1}{2^2-1} = \frac{1}{3} = \frac{1}{2\times1+1} \Rightarrow H_1$ is true | B1 | |
| $\frac{k}{2k+1} + \frac{1}{(2k+2)^2-1} = \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} = \frac{2k^2+3k+1}{(2k+1)(2k+3)}$ | M1A1 | |
| $= \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2[k+1]+1}$ | A1 | |
| $\therefore H_k \Rightarrow H_{k+1}$; by PMI $H_n$ true for all positive integers $n$ | A1 | Requires all previous marks |
| $\sum_{r=1}^{\infty} \frac{1}{(2r)^2-1} = \frac{1}{2}$ | B1 | |
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3 Prove by mathematical induction that, for all positive integers $n , \sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }$.
State the value of $\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( 2 r ) ^ { 2 } - 1 }$.
\hfill \mbox{\textit{CAIE FP1 2015 Q3 [7]}}