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Summation by Sigma Method

Prove that ...

Question

Prove that $n \sum k = 1 k 2^{- k} = 2 [1 - 2^{- n} - n {.2}^{- (n + 1)}]$ .

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Solution

$S_{n} = {1.2}^{- 1} + {2.2}^{- 2} + {3.2}^{- 3} + . . . . . . . . + n {.2}^{- n}$
$2^{- 1} S_{n} = 2^{- 2} + {2.2}^{- 3} + . . . . . . . . + (n - 1) {.2}^{- n} + n {.2}^{- (n + 1)}$
$(1 - 2^{- 1}) S_{n} = 2^{- 1} + 2^{- 2} + 2^{- 3} + . . . . . . + 2^{- n} - n {.2}^{- (n + 1)}$
$2^{- 1} S_{n} = 2^{- 1} \frac{2^{- n} - 1}{2^{- 1} - 1} - n {.2}^{- (n + 1)}$
$S_{n} = 2 (1 - 2^{- n} - n {.2}^{- (n + 1)})$

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