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Question Number 136723 by mnjuly1970 last updated on 25/Mar/21

$$....calculus\left(I\right).....$$$$provethat::$$$$f\left(x\right)=\frac{1}{\sqrt{1-4x}}\stackrel{???}{=}\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}2n\\ n\end{array}\right)x^n$$$$$$

Answered by Dwaipayan Shikari last updated on 25/Mar/21

$$\frac{1}{\sqrt{1-4x}}=1+\frac{4x}{2.1!}+\frac{{\left(4x\right)}^2}{2^2.2!}+\frac{{\left(4x\right)}^3}{2^3.3!}+\frac{{\left(4x\right)}^4}{2^4.4!}+...$$$$=1+\left(\begin{array}{c}2\\ 1\end{array}\right)x^1+\left(\begin{array}{c}4\\ 2\end{array}\right)x^2+\left(\begin{array}{c}6\\ 3\end{array}\right)x^3+...=\underset{n=0}{\sum ^{\mathrm{\infty }}}\left(\begin{array}{c}2n\\ n\end{array}\right)x^n$$$$Convergeswhenx<\frac{1}{4}$$$$\sqrt{5}=1+\frac{4}{2.5.1!}+\frac{4^2}{5^2.2^2.2!}+..adinf$$

Commented by mnjuly1970 last updated on 25/Mar/21

$$thanksalot...$$

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