3 Let \(S _ { N } = \frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { N } { ( N + 1 ) ! }\). Prove by mathematical induction that, for all positive integers \(N\),
$$S _ { N } = 1 - \frac { 1 } { ( N + 1 ) ! }$$
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Question 3:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(H_N: S_N = 1 - \frac{1}{(N+1)!}\) —
Proposition stated
\(S_1 = \frac{1}{2!} = \frac{1}{2} = 1 - \frac{1}{2!} \Rightarrow H_1\) is true B1
Proves base case
\(H_k\): Assume \(S_k = 1 - \frac{1}{(k+1)!}\) is true B1
States inductive hypothesis
\(\Rightarrow S_{k+1} = 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{(k+2)!-(k+2)+(k+1)}{(k+2)!}\) M1
Proves inductive step
\(\Rightarrow S_{k+1} = 1 - \frac{1}{(k+2)!}\) \(\therefore H_k \Rightarrow H_{k+1}\) A1
\(\therefore\) By PMI \(H_n\) is true for all positive integers \(N\) A1
States conclusion
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## Question 3:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $H_N: S_N = 1 - \frac{1}{(N+1)!}$ | — | Proposition stated |
| $S_1 = \frac{1}{2!} = \frac{1}{2} = 1 - \frac{1}{2!} \Rightarrow H_1$ is true | B1 | Proves base case |
| $H_k$: Assume $S_k = 1 - \frac{1}{(k+1)!}$ is true | B1 | States inductive hypothesis |
| $\Rightarrow S_{k+1} = 1 - \frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = \frac{(k+2)!-(k+2)+(k+1)}{(k+2)!}$ | M1 | Proves inductive step |
| $\Rightarrow S_{k+1} = 1 - \frac{1}{(k+2)!}$ $\therefore H_k \Rightarrow H_{k+1}$ | A1 | |
| $\therefore$ By PMI $H_n$ is true for all positive integers $N$ | A1 | States conclusion |
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3 Let $S _ { N } = \frac { 1 } { 2 ! } + \frac { 2 } { 3 ! } + \frac { 3 } { 4 ! } + \ldots + \frac { N } { ( N + 1 ) ! }$. Prove by mathematical induction that, for all positive integers $N$,
$$S _ { N } = 1 - \frac { 1 } { ( N + 1 ) ! }$$
\hfill \mbox{\textit{CAIE FP1 2012 Q3 [5]}}