Sum of Sequence of n by 2 to the Power of n/Proof 1

Theorem

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

Proof

From Sum of Arithmetic-Geometric Sequence:

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

Hence:

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

\(=\)

\(\ds \frac {0 \paren {1 - 2^{n + 1} } } {1 - 2} + \frac {2 \times 1 \paren {1 - \paren {n + 1} 2^n + n 2^{n + 1} } } {\paren {1 - 2}^2}\)

putting $a = 0, d = 1, r = 2$

\(\ds \)

\(=\)

\(\ds 2 \paren {1 - \paren {n + 1} 2^n + n 2^{n + 1} }\)

initial simplification

\(\ds \)

\(=\)

\(\ds 2 - \paren {n + 1} 2^{n + 1} + n 2^{n + 2}\)

further simplification

Hence the result.

$\blacksquare$