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Question Number 162249 by SANOGO last updated on 28/Dec/21

Answered by mr W last updated on 28/Dec/21

$$\left(1\right)$$$$\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)x^k={(1+x)}^n$$$$setx=-1$$$$\Rightarrow \underset{k=0}{\sum ^n}{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)=0$$$$\underset{n\to \mathrm{\infty }}{\lim }\underset{k=0}{\sum ^n}{(-1)}^k\left(\begin{array}{c}n\\ k\end{array}\right)=\underset{n\to \mathrm{\infty }}{\lim 0}=0$$$$$$$$\left(2\right)$$$$\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)x^k={(1+x)}^n$$$$\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right){\int }_0^x{x}^{k}dx={\int }_0^x{(1+x)}^{n}dx$$$$\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{x^{k+1}}{k+1}={\left[\frac{{(1+x)}^{n+1}}{n+1}\right]}_0^x$$$$\underset{k=0}{\sum ^n}\left(\begin{array}{c}n\\ k\end{array}\right)\frac{x^{k+1}}{k+1}=\frac{{(1+x)}^{n+1}-1}{n+1}$$$$setx=1$$$$\Rightarrow \underset{k=0}{\sum ^n}\frac{1}{k+1}\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{2^{n+1}-1}{n+1}$$

Commented by SANOGO last updated on 28/Dec/21

$$mercibien$$

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