Question Number 136723 by mnjuly1970 last updated on 25/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{calculus}\:\:\:\:\:\left({I}\right)….. \\ $$$$\:\:\:\:\:{prove}\:\:\:{that}\::: \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}\:\overset{???} {=}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\:\mathrm{2}{n}}\\{\:\:\:{n}}\end{pmatrix}\:{x}^{{n}} \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 25/Mar/21
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}=\mathrm{1}+\frac{\mathrm{4}{x}}{\mathrm{2}.\mathrm{1}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} .\mathrm{4}!}+… \\ $$$$=\mathrm{1}+\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}{x}^{\mathrm{1}} +\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}{x}^{\mathrm{3}} +…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}{x}^{{n}} \\ $$$${Converges}\:{when}\:{x}<\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sqrt{\mathrm{5}}=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{2}.\mathrm{5}.\mathrm{1}!}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} .\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+..{ad}\:{inf} \\ $$
Commented by mnjuly1970 last updated on 25/Mar/21
$$\:\:{thanks}\:{alot}\:… \\ $$