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Question Number 205671 by hardmath last updated on 26/Mar/24

Answered by Berbere last updated on 27/Mar/24

$${(1-x)}^n=\underset{k=0}{\sum ^n}{(-1)}^k{x}^k\left(\begin{array}{c}n\\ k\end{array}\right)\Rightarrow {\int }_0^t\underset{k=0}{\sum ^n}{(-1)}^k{x}^k\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{1}{n+1}-\frac{{(1-x)}^{n+1}}{n+1}$$$$\Rightarrow \underset{k=0}{\sum ^n}\frac{{(-1)}^k{x}^{k+1}}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{1}{n+1}(1-{(1-x)}^{n+1})$$$$\frac{n+2}{2^n}\underset{k=0}{\sum ^n}\frac{{(-1)}^k{2}^{n-k}}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=(n+2)\underset{k=0}{\sum ^n}\frac{{(-1)}^k{\left(\frac{1}{2}\right)}^k}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)$$$$=2(n+2)\underset{k=0}{\sum ^n}\frac{{(-1)}^k{\left(\frac{1}{2}\right)}^{k+1}}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=2(n+2).\frac{1}{n+1}(1-{(1-\frac{1}{2})}^{n+1})$$$$\to 2$$

Commented by hardmath last updated on 27/Mar/24

$$\mathrm{Thank}\mathrm{you}\mathrm{my}\mathrm{dear}\mathrm{professor}\mathrm{perfect}$$

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