6 Prove by induction that
$$\frac { 2 \times 1 } { 2 \times 3 } + \frac { 2 ^ { 2 } \times 2 } { 3 \times 4 } + \frac { 2 ^ { 3 } \times 3 } { 4 \times 5 } + \ldots + \frac { 2 ^ { n } \times n } { ( n + 1 ) ( n + 2 ) } = \frac { 2 ^ { n + 1 } } { n + 2 } - 1$$
for all integers \(n \geqslant 1\).
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- Assume result true for \(n = k\) - Then \(\sum_{r=1}^{k+1} \frac{2^r \times r}{(r+1)(r+2)} = \frac{2^{k+1}}{k+2} + \frac{2^{k+1}(k+1)}{(k+2)(k+3)} - 1\) M1A1
- \(= \frac{2^{k+1}(k+3+k+1)}{(k+2)(k+3)} - 1\) M1
Putting over common denominator (not including the \(-1\), unless separated later)
- \(= \frac{2^{k+1} \cdot 2(k+2)}{(k+2)(k+3)} - 1\) A1
- \(= \frac{2^{k+2}}{k+3} - 1\) A1
- \(k = 1: \text{LHS} = \frac{1}{3}, \text{RHS} = \frac{2^2}{3} - 1\) B1
7 marks; Must be completely correct
- \(P_k \Rightarrow P_{k+1}\) and \(P_1\) true E1
- Total: 7 marks
Question 6 Total: 7 marks
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- Assume result true for $n = k$ | |
- Then $\sum_{r=1}^{k+1} \frac{2^r \times r}{(r+1)(r+2)} = \frac{2^{k+1}}{k+2} + \frac{2^{k+1}(k+1)}{(k+2)(k+3)} - 1$ | M1A1 |
- $= \frac{2^{k+1}(k+3+k+1)}{(k+2)(k+3)} - 1$ | M1 | Putting over common denominator (not including the $-1$, unless separated later)
- $= \frac{2^{k+1} \cdot 2(k+2)}{(k+2)(k+3)} - 1$ | A1 |
- $= \frac{2^{k+2}}{k+3} - 1$ | A1 |
- $k = 1: \text{LHS} = \frac{1}{3}, \text{RHS} = \frac{2^2}{3} - 1$ | B1 | 7 marks; Must be completely correct
- $P_k \Rightarrow P_{k+1}$ and $P_1$ true | E1 |
- **Total: 7 marks**
**Question 6 Total: 7 marks**
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6 Prove by induction that
$$\frac { 2 \times 1 } { 2 \times 3 } + \frac { 2 ^ { 2 } \times 2 } { 3 \times 4 } + \frac { 2 ^ { 3 } \times 3 } { 4 \times 5 } + \ldots + \frac { 2 ^ { n } \times n } { ( n + 1 ) ( n + 2 ) } = \frac { 2 ^ { n + 1 } } { n + 2 } - 1$$
for all integers $n \geqslant 1$.
\hfill \mbox{\textit{AQA FP2 2009 Q6 [7]}}