Sum of Sequence of Power by Index/Proof 2

Proof
From Sum of Arithmetic-Geometric Progression:


 * $\displaystyle \sum_{j \mathop = 0}^n \paren {a + j d} x^j = \frac {a \paren {1 - x^{n + 1} } } {1 - x} + \frac {x d \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}$

Hence:

Hence the result.