Question Number 11648 by uni last updated on 29/Mar/17
$$\mid\mathrm{x}\mid<\mathrm{l} \\ $$ $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{x}^{\mathrm{n}} =? \\ $$
Answered by Nayon last updated on 29/Mar/17
$$ \\ $$ $$ \\ $$ $$={x}^{\mathrm{1}} +{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +.......... \\ $$ $${let}\:{p}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +..... \\ $$ $$=>{px}={x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +......... \\ $$ $$=>{p}−{px}={x} \\ $$ $$=>{p}\left(\mathrm{1}−{x}\right)={x} \\ $$ $$=>{p}=\frac{{x}}{\mathrm{1}−{x}}\: \\ $$ $$ \\ $$