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Question Number 151181 by mathdanisur last updated on 18/Aug/21

$$\mathrm{if}\mid \mathbf{x}\mid <1$$$$\mathrm{find}\mathrm{x}-{4\mathrm{x}}^2+{9\mathrm{x}}^3-{16\mathrm{x}}^4+...$$

Answered by Olaf_Thorendsen last updated on 18/Aug/21

$$\mathrm{S}\left(x\right)=-\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{n}^2{x}^n$$$$\mathrm{Let}f\left(x\right)=\frac{1}{1+x}=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n{x}^n$$$$f^{\prime }\left(x\right)=-\frac{1}{{(1+x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{nx}^{n-1}$$$${xf}^{\prime }\left(x\right)=-\frac{x}{{(1+x)}^2}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{nx}^n$$$$-\frac{1-x}{{(1+x)}^3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{n}^2{x}^{n-1}$$$$\frac{x(1-x)}{{(1+x)}^3}=-\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{n}^2{x}^n=\mathrm{S}\left(x\right)$$

Commented by mathdanisur last updated on 18/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

Commented by mathdanisur last updated on 19/Aug/21

$$\mathrm{Dear}\mathbf{S}\mathrm{er},\mathrm{finally}\mathrm{answer}\frac{\mathrm{x}(1-\mathrm{x})}{{(1+\mathrm{x})}^3}$$$$\mathrm{or}\frac{{2\mathrm{x}}^2}{{(1+\mathrm{x})}^3}.?$$

Answered by qaz last updated on 19/Aug/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{n}}^2{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}(\mathrm{n}(\mathrm{n}+1)-\mathrm{n}){(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\mathrm{n}(\mathrm{n}+1){(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\mathrm{n}{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}$$$$=2\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\mathrm{k}{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}$$$$=2\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\mathrm{k}{(-1)}^{\mathrm{n}+\mathrm{k}-1}{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}-\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}+\mathrm{k}-1}{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}$$$$=\frac{2}{1+\mathrm{x}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{i}=0}{\sum ^{\mathrm{k}}}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}+1}-\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}$$$$=\frac{2}{1+\mathrm{x}}\underset{\mathrm{i}=0}{\sum ^{\mathrm{\infty }}}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{k}+\mathrm{i}}{\mathrm{x}}^{\mathrm{k}+\mathrm{i}+1}-\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}$$$$=\frac{2\mathrm{x}}{{(1+\mathrm{x})}^3}-\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}$$$$=\frac{\mathrm{x}-{\mathrm{x}}^2}{{(1+\mathrm{x})}^3}$$$$----------------$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{n}}^2{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}({\mathrm{n}}^2+2\mathrm{n}+1){(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}$$$$=\mathrm{x}({\left(\mathrm{yD}\right)}^2+2\mathrm{yD}+1){\mid }_{\mathrm{y}=-\mathrm{x}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{y}}^{\mathrm{n}}$$$$=\mathrm{x}({\mathrm{y}}^2{\mathrm{D}}^2+3\mathrm{yD}+1){\mid }_{\mathrm{y}=-\mathrm{x}}\frac{1}{1-\mathrm{y}}$$$$=\mathrm{x}{[\frac{{2\mathrm{y}}^2}{{(1-\mathrm{y})}^3}+\frac{3\mathrm{y}}{{(1-\mathrm{y})}^2}+\frac{1}{1-\mathrm{y}}]}_{\mathrm{y}=-\mathrm{x}}$$$$=\frac{\mathrm{x}(1-\mathrm{x})}{{(1+\mathrm{x})}^3}$$

Commented by mathdanisur last updated on 19/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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