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

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Theorem

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


Proof 1

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.

$\blacksquare$


Proof 2

From Sum of Sequence of Power by Index:

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


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.

$\blacksquare$


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