Sum of Special Series involving exponents

Question

What is the sum to $n$ terms of the series $$2(2^0)+ 3(2^1) + 4(2^2) + \cdots$$

My try:- The $n$th term is $$(n+1)(2^{n-1}) = n(2^{n-1}) + 2^{n-1}.$$ So the sum is the summation of these two terms. I know the summation of the second term using G.P. sum but I do not know the summation of the first term as it involves multiplication.

I would also like to know other ways of finding the sum.

Sorry for the poor presentation, I do not know how to use Latex.

Thank you.

Answer 1

Hint

Consider $$\sum_{n=0}^p (n+1)x^n=\sum_{n=0}^p nx^n+\sum_{n=0}^p x^n=x\sum_{n=0}^p nx^{n-1}+\sum_{n=0}^p x^n$$

When done, make $x=2$

Answer 2

It is an AGP. So,
$S_n=2.2^0+3.2^1+4.2^2+ ........ +(n+1).2^{n-1}$

Multiply the equation by 2

$S_{n}=2.2^0+3.2^1+4.2^2+ ........ +(n+1).2^{n-1}+0$ $2.S_n=0+2.2^1+3.2^2+4.2^3+ ..... +n.2^{n-1} +(n+1).2^{n}$ Now subtracting $S_n=-2-(2^1+2^2+....+2^{n-1})+(n+1).2^n$

$S_n $ can be easily found from here