Sum of Sequence of n by 2 to the Power of n/Proof 2

Theorem

$\ds \sum_{j \mathop = 0}^n j \, 2^j = n \, 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$

$\ds \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$

Hence:

$\ds \ds \sum_{j \mathop = 0}^n j \, 2^j$

$=$

$\ds \frac {n 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2} {\paren {2 - 1}^2}$

putting $x = 2$

$\ds $

$=$

$\ds n 2^{n + 2} - \paren {n + 1} 2^{n + 1} + 2$

simplification

Hence the result.

$\blacksquare$