How to get this required RHS from the given LHS. $$\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$$
This was used in a solution to the following question I asked. I couldn't understand the step and hence to understand in detail I put it as a question.
Find the sum of the n terms of the series $2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$
$$ \sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{2^k{x}^{k}}\\ \sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{{(2x)}^{k}}\\ \sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\frac{1-{(2x)}^{n+1}}{1-2x}\\ \sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-x^2{2}^{(n+1)}{x}^{n+1}}{1-2x}\\ \sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-{2}^{n+1}{x}^{n+3}}{1-2x} $$
I used the formula for summing geometric series:
$$ \sum_{k=0}^{n}{t^k} = \frac{1-{t}^{n+1}}{1-t} $$
Use the geometric sum formula: $$\sum_{k=0}^n2^kx^{k+2}=x^2\sum_{k=0}^n(2x)^{k} =x^2\frac{1-(2x)^{n+1}}{1-2x}$$
