Answer
See explanations.
Work Step by Step
Step 1. Prove the formula is true when $n=1$: $LHS=1\times2=2$, $RHS=2(1+(1-1)2^1)=2$, thus it is true for $n=1$.
Step 2. Assume the formula is true when $n=k$: we have $1\times2+2\times2^2+3\times2^3+...+2^kk=2[1+(k-1)2^k]$
Step 3. Prove it is true for $n=k+1$: $LHS=1\times2+2\times2^2+3\times2^3+...+2^kk+2^{k+1}(k+1)
=2[1+(k-1)2^k]+2^{k+1}(k+1)=2+(k-1)2^{k+1}+(k+1)2^{k+1}=2+2^{k+2}k$
Use $n=k+1$ on the right side of the formula, we have $RHS=2[1+(k)2^{k+1}]=2+2^{k+2}k$
Thus, we have $LHS=RHS$ which proves the formula is also true for $n=k+1$
Step 4. Conclusion: with mathematical induction, we have proved that the formula is true for all natural numbers n.
