Question Number 181207 by lapache last updated on 23/Nov/22
$${cacul} \\ $$$$\left.\forall{x}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\underset{{n}=\mathrm{0}} {\overset{+\infty} {\sum}}{nx}^{{n}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 23/Nov/22
$${S}={x}+\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$${xS}={x}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$${S}−{xS}={x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +\centerdot\centerdot\centerdot=\frac{{x}}{\mathrm{1}−{x}} \\ $$$$\left(\mathrm{1}−{x}\right){S}=\frac{{x}}{\mathrm{1}−{x}}\:\Rightarrow{S}=\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$