2 Prove by mathematical induction that, for all positive integers \(n\),
$$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \frac { 1 } { 2 x + 3 } \right) = ( - 1 ) ^ { n } \frac { n ! 2 ^ { n } } { ( 2 x + 3 ) ^ { n + 1 } }$$
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Question 2:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\(P_n: \frac{d^n}{dx^n}\left(\frac{1}{2x+3}\right) = (-1)^n \frac{n!\,2^n}{(2x+3)^{n+1}}\) States proposition
\(\frac{d}{dx}\left(\frac{1}{2x+3}\right) = (-1)(2x+3)^{-2}\times 2\) M1
Proves base case
\(= (-1)\frac{1!\times 2}{(2x+3)^2} \Rightarrow P_1\) is true A1
Assume \(P_k\) is true: \(\frac{d^k}{dx^k}\left(\frac{1}{2x+3}\right) = (-1)^k \frac{k!\,2^k}{(2x+3)^{k+1}}\) B1
States inductive hypothesis
\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right) = (-1)^{k+1}\frac{2(k+1)k!\,2^k}{(2x+3)^{k+2}}\) M1
Shows \(P_k \Rightarrow P_{k+1}\)
\(= (-1)^{k+1}\frac{(k+1)!\,2^{k+1}}{(2x+3)^{k+2}}\) A1
Since \(P_1\) is true and \(P_k \Rightarrow P_{k+1}\), by mathematical induction \(P_n\) is true \(\forall n \in \mathbb{Z}^+\) A1
States conclusion; Part marks: 6; Total: [6]
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## Question 2:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $P_n: \frac{d^n}{dx^n}\left(\frac{1}{2x+3}\right) = (-1)^n \frac{n!\,2^n}{(2x+3)^{n+1}}$ | | States proposition |
| $\frac{d}{dx}\left(\frac{1}{2x+3}\right) = (-1)(2x+3)^{-2}\times 2$ | M1 | Proves base case |
| $= (-1)\frac{1!\times 2}{(2x+3)^2} \Rightarrow P_1$ is true | A1 | |
| Assume $P_k$ is true: $\frac{d^k}{dx^k}\left(\frac{1}{2x+3}\right) = (-1)^k \frac{k!\,2^k}{(2x+3)^{k+1}}$ | B1 | States inductive hypothesis |
| $\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right) = (-1)^{k+1}\frac{2(k+1)k!\,2^k}{(2x+3)^{k+2}}$ | M1 | Shows $P_k \Rightarrow P_{k+1}$ |
| $= (-1)^{k+1}\frac{(k+1)!\,2^{k+1}}{(2x+3)^{k+2}}$ | A1 | |
| Since $P_1$ is true and $P_k \Rightarrow P_{k+1}$, by mathematical induction $P_n$ is true $\forall n \in \mathbb{Z}^+$ | A1 | States conclusion; Part marks: 6; **Total: [6]** |
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2 Prove by mathematical induction that, for all positive integers $n$,
$$\frac { \mathrm { d } ^ { n } } { \mathrm {~d} x ^ { n } } \left( \frac { 1 } { 2 x + 3 } \right) = ( - 1 ) ^ { n } \frac { n ! 2 ^ { n } } { ( 2 x + 3 ) ^ { n + 1 } }$$
\hfill \mbox{\textit{CAIE FP1 2011 Q2 [6]}}