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

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

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

$$S=\underset{k=0}{\sum ^n}k{2}^k$$$$2S=\underset{k=0}{\sum ^n}(k+1-1)2^{k+1}$$$$2S=\underset{k=0}{\sum ^n}(k+1)2^{k+1}-\underset{k=0}{\sum ^n}{2}^{k+1}$$$$2S=\underset{k=0}{\sum ^{n+1}}k{2}^k-\underset{k=0}{\sum ^n}{2}^{k+1}$$$$2S=(n+1)2^{n+1}+S-\underset{k=0}{\sum ^n}{2}^{k+1}$$$$S=(n+1)2^{n+1}-\underset{k=0}{\sum ^n}{2}^{k+1}$$$$S=(n+1)2^{n+1}-\frac{2(2^{n+1}-1)}{2-1}$$$$\Rightarrow S=(n-1)2^{n+1}+2$$$$$$$$generally:$$$$\underset{k=0}{\sum ^n}{kp}^k=\frac{1}{p-1}(n-\frac{1}{p-1})p^{n+1}+\frac{p}{{(p-1)}^2}$$$$withp\ne 1$$

Commented by Abdoulaye last updated on 01/Mar/21

$$thankyousir$$

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

$$\underset{k=0}{\sum ^n}k{2}^k$$$$={\left[xD\underset{k=0}{\sum ^n}{x}^k\right]}_{x=2}$$$$=xD{\left[\frac{x^{n+1}-1}{x-1}\right]}_{x=2}$$$$=\frac{{nx}^{n+2}-(n+1)x^{n+1}+x}{{(x-1)}^2}{\mid }_{x=2}$$$$=n{2}^{n+2}-(n+1)2^{n+1}+2$$$$=(n-1)2^{n+1}+2$$

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