Prove the closed form of a summation by induction

Question

Prove by induction that for every positive integer $n$:

$$\sum_{j=1}^nj2^j = (n − 1)2^{n+1}+ 2.$$

Answer 1

For $n=1$, $LHS=1(2^1)=2, RHS=(1-1)2^{1+1}+2=2$. The statement is true for $n=1$.

Then for the induction step, if $n=k$ is true, then when $n=k+1$:

$LHS=\sum_{j=1}^{k+1} j2^j=(k-1)2^{k+1}+2+(k+1)2^{k+1}=(2k)2^{k+1}+2=k2^{k+2}+2 \\ RHS=(k+1-1)2^{k+1+1}+2=k2^{k+2}+2$

The statement is true for $n=k+1$.

Therefore, by M.I., the sum is true.