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Question Number 134200 by Abdoulaye last updated on 28/Feb/21

h o w d o w e c a l c u l a t e

\underset{k = 0}{\sum^{n}} k 2^{k} = ?

Answered by mr W last updated on 01/Mar/21

S = \underset{k = 0}{\sum^{n}} k 2^{k}

2 S = \underset{k = 0}{\sum^{n}} (k + 1 - 1) 2^{k + 1}

2 S = \underset{k = 0}{\sum^{n}} (k + 1) 2^{k + 1} - \underset{k = 0}{\sum^{n}} 2^{k + 1}

2 S = \underset{k = 0}{\sum^{n + 1}} k 2^{k} - \underset{k = 0}{\sum^{n}} 2^{k + 1}

2 S = (n + 1) 2^{n + 1} + S - \underset{k = 0}{\sum^{n}} 2^{k + 1}

S = (n + 1) 2^{n + 1} - \underset{k = 0}{\sum^{n}} 2^{k + 1}

S = (n + 1) 2^{n + 1} - \frac{2 (2^{n + 1} - 1)}{2 - 1}

\Rightarrow S = (n - 1) 2^{n + 1} + 2

g e n e r a l l y :

\underset{k = 0}{\sum^{n}} {k p}^{k} = \frac{1}{p - 1} (n - \frac{1}{p - 1}) p^{n + 1} + \frac{p}{{(p - 1)}^{2}}

w i t h p \neq 1

Commented by Abdoulaye last updated on 01/Mar/21

t h a n k y o u s i r

Answered by Ñï= last updated on 01/Mar/21

\underset{k = 0}{\sum^{n}} k 2^{k}

= {[x D \underset{k = 0}{\sum^{n}} x^{k}]}_{x = 2}

= x D {[\frac{x^{n + 1} - 1}{x - 1}]}_{x = 2}

= \frac{{n x}^{n + 2} - (n + 1) x^{n + 1} + x}{{(x - 1)}^{2}} ∣_{x = 2}

= n 2^{n + 2} - (n + 1) 2^{n + 1} + 2

= (n - 1) 2^{n + 1} + 2

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