Answer
See below.
Work Step by Step
Proofs using mathematical induction consist of two steps:
1) The base case: here we prove that the statement holds for the first natural number.
2) The inductive step: assume the statement is true for some arbitrary natural number $n$ larger than the first natural number. Then we prove that then the statement also holds for $n + 1$.
Hence, here we have:
1) For $n=1: 1(2)=2(1+(1-1)2^1)$.
2) Assume for $n=k: 1(2)+2(2)^2++k(2^k)= 2(1+(k-1)2^k)$. Then for $n=k+1$:
$1(2)+2(2)^2++k(2^k)+(k+1)(2^{k+1})= 2(1+(k-1)2^k)+(k+1)(2^{k+1})=2(1+(k)2^{k+1})$
Thus we proved what we wanted to.