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Question Number 197639 by mokys last updated on 25/Sep/23

$$showthat\frac{1}{{(1-z)}^n}=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{(n+k-1)!}{k!(n-1)!}{z}^k$$

Commented by mr W last updated on 25/Sep/23

$$\frac{1}{{(1-z)}^n}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}k+n-1\\ k\end{array}\right)z^k=\underset{k=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}k+n-1\\ n-1\end{array}\right)z^k$$

Answered by mr W last updated on 25/Sep/23

$$f\left(x\right)=\frac{1}{{(1-x)}^n}={(1-x)}^{-n}$$$$f\left(0\right)=1$$$$f^{\left(1\right)}\left(x\right)=n{(1-x)}^{-(n+1)}\Rightarrow f^{\left(1\right)}\left(0\right)=n$$$$f^{\left(2\right)}\left(x\right)=n(n+1){(1-x)}^{-(n+2)}\Rightarrow f^{\left(2\right)}\left(0\right)=n(n+1)$$$$...$$$$similarly$$$$f^{\left(k\right)}\left(0\right)=n(n+1)...(n+k-1)$$$$acc.totaylor$$$$f\left(x\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{n(n+1)...(n+k-1)}{k!}{x}^k$$$$=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{(n+k-1)!}{k!(n-1)!}{x}^k✓$$

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