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Question Number 123461 by Algoritm last updated on 25/Nov/20

Answered by Snail last updated on 25/Nov/20

$$f\left(x\right)={(1-x)}^{-1}$$$$f\left(1/4\right)=4/3$$

Answered by Olaf last updated on 25/Nov/20

$$\mathrm{Let}{\mathrm{P}}_n\left(x\right)=\underset{k=0}{\prod ^{n-1}}(x+k),n⩾1$$$${\mathrm{P}}_n\left(\frac{1}{4}\right)=\underset{k=0}{\prod ^{n-1}}(\frac{1}{4}+k)$$$${\mathrm{P}}_n\left(\frac{1}{4}\right)=\underset{k=0}{\prod ^{n-1}}\frac{k+4}{4}$$$${\mathrm{P}}_n\left(\frac{1}{4}\right)=\frac{(n+3)!}{4^{n}3!}$$$$f\left(x\right)=1+\underset{n=1(}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{P}}_n\left(x\right)}{n+1)!}$$$$f\left(\frac{1}{4}\right)=1+\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n+3)!}{4^n(n+1)!}$$$$f\left(\frac{1}{4}\right)=1+\frac{1}{6}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{(n+2)(n+3)}{4^n}$$$$f\left(\frac{1}{4}\right)=\frac{3}{2}+\frac{8}{3}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n(n+1)}{4^n}$$$$\mathrm{Let}g\left(x\right)=\frac{1}{1-x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n$$$$g^{\prime }\left(x\right)=\frac{1}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n-1}$$$$x^2{g}^{\prime }\left(x\right)=\frac{x^2}{{(1-x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n+1}$$$$\frac{2x{(1-x)}^2+2x^2(1-x)}{{(1-x)}^4}=\underset{n=1}{\sum ^{\mathrm{\infty }}}n(n+1)x^n$$$$\mathrm{If}x=\frac{1}{4}:\frac{32}{27}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{n(n+1)}{4^n}$$$$f\left(\frac{1}{4}\right)=\frac{3}{2}+\frac{8}{3}\times \frac{32}{27}=\frac{755}{162}$$

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