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

               ....calculus     (I).....       prove   that ::      f(x)= (1/( (√(1−4x)))) =^(???) Σ_(n=0) ^∞  ((( 2n)),((   n)) ) x^n

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{calculus}\:\:\:\:\:\left({I}\right)..... \\ $$$$\:\:\:\:\:{prove}\:\:\:{that}\::: \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}\:\overset{???} {=}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\:\mathrm{2}{n}}\\{\:\:\:{n}}\end{pmatrix}\:{x}^{{n}} \\ $$$$ \\ $$

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

(1/( (√(1−4x))))=1+((4x)/(2.1!))+(((4x)^2 )/(2^2 .2!))+(((4x)^3 )/(2^3 .3!))+(((4x)^4 )/(2^4 .4!))+...  =1+ ((2),(1) )x^1 + ((4),(2) )x^2 + ((6),(3) )x^3 +...=Σ_(n=0) ^∞  (((2n)),(n) )x^n   Converges when x<(1/4)  (√5)=1+(4/(2.5.1!))+(4^2 /(5^2 .2^2 .2!))+..ad inf

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{4}{x}}}=\mathrm{1}+\frac{\mathrm{4}{x}}{\mathrm{2}.\mathrm{1}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} .\mathrm{3}!}+\frac{\left(\mathrm{4}{x}\right)^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} .\mathrm{4}!}+... \\ $$$$=\mathrm{1}+\begin{pmatrix}{\mathrm{2}}\\{\mathrm{1}}\end{pmatrix}{x}^{\mathrm{1}} +\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}{x}^{\mathrm{2}} +\begin{pmatrix}{\mathrm{6}}\\{\mathrm{3}}\end{pmatrix}{x}^{\mathrm{3}} +...=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}{x}^{{n}} \\ $$$${Converges}\:{when}\:{x}<\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\sqrt{\mathrm{5}}=\mathrm{1}+\frac{\mathrm{4}}{\mathrm{2}.\mathrm{5}.\mathrm{1}!}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{5}^{\mathrm{2}} .\mathrm{2}^{\mathrm{2}} .\mathrm{2}!}+..{ad}\:{inf} \\ $$

Commented by mnjuly1970 last updated on 25/Mar/21

  thanks alot ...

$$\:\:{thanks}\:{alot}\:... \\ $$

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