Q12 (b)

Criteria	Marks
Provides correct solution	3
Establishes the inductive step, or equivalent merit	2
Establishes the base case, or equivalent merit	1

Sample answer:

When $n=1$,

$$\begin{aligned} \text{LHS}&=1\times2=2\\ \text{RHS}&=2+(1-1)2^{1+1}=2\\ \text{LHS}&=\text{RHS} \end{aligned}$$

Therefore, true when $n=1$.

Assume true when $n=k$, i.e.

$$(1\times2)+(2\times2^2)+\cdots+(k\times2^k)=2+(k-1)2^{k+1}$$

Consider $n=k+1$.

$$\begin{aligned} \text{RHS}&=2+(k+1-1)2^{k+1+1}=2+k\cdot2^{k+2}\\ \text{LHS}&=(1\times2)+(2\times2^2)+\cdots+(k\times2^k)+((k+1)\times2^{k+1})\\ &=2+(k-1)2^{k+1}+(k+1)2^{k+1}\quad\text{by assumption}\\ &=2+2^{k+1}(k-1+k+1)\\ &=2+2^{k+1}(2k)\\ &=2+k\cdot2^{k+2}\\ &=\text{RHS} \end{aligned}$$

Therefore, by mathematical induction, it is true for all integers $n\ge1$.