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Question Number 177880 by DAVONG last updated on 10/Oct/22

Answered by mr W last updated on 10/Oct/22

$${(1+x)}^n=\underset{k=0}{\sum ^n}{C}_k^n{x}^k$$$$n{(1+x)}^{n-1}=\underset{k=1}{\sum ^n}{kC}_k^n{x}^{k-1}$$$$withx=1:$$$$n{(1+1)}^{n-1}=\underset{k=1}{\sum ^n}{kC}_k^n=11264=11\times 2^{10}$$$$n{2}^{n-1}=11\times 2^{11-1}$$$$\Rightarrow n=11$$

Commented by DAVONG last updated on 10/Oct/22

$$\mathrm{Thanks}\mathrm{teacher}$$

Commented by Tawa11 last updated on 10/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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