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Question Number 98205 by I want to learn more last updated on 12/Jun/20

$$\mathbf{Find}\mathbf{the}\mathbf{nth}\mathbf{term}\mathbf{of}\mathbf{the}\mathbf{sequence}\left\{{\mathbf{a}}_{\mathbf{n}}\right\}\mathbf{such}\mathbf{that}$$$${\mathbf{a}}_1=1,{\mathbf{a}}_{\mathbf{n}+1}=\frac{1}{2}{\mathbf{a}}_{\mathbf{n}}+\frac{{\mathbf{n}}^2-2\mathbf{n}-1}{{\mathbf{n}}^2{(\mathbf{n}+1)}^2}(\mathbf{n}=1,2,3,...)$$

Answered by mr W last updated on 12/Jun/20

$$\frac{n^2-2n-1}{n^2{(n+1)}^2}$$$$=\frac{2n^2-{(n+1)}^2}{n^2{(n+1)}^2}$$$$=\frac{2}{{(n+1)}^2}-\frac{1}{n^2}$$$$\Rightarrow a_{n+1}=\frac{1}{2}{a}_n+\frac{2}{{(n+1)}^2}-\frac{1}{n^2}$$$$\Rightarrow a_{n+1}-\frac{2}{{(n+1)}^2}=\frac{1}{2}(a_n-\frac{2}{n^2})$$$$let{b}_n=a_n-\frac{2}{n^2}$$$$\Rightarrow b_{n+1}=\frac{1}{2}{b}_n\Rightarrow G.P.$$$$\Rightarrow b_n=b_1{\left(\frac{1}{2}\right)}^{n-1}=(a_1-\frac{2}{1^1})\frac{1}{2^{n-1}}=-\frac{1}{2^{n-1}}$$$$\Rightarrow a_n=-\frac{1}{2^{n-1}}+\frac{2}{n^2}=2(\frac{1}{n^2}-\frac{1}{2^n})$$

Commented by I want to learn more last updated on 12/Jun/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by I want to learn more last updated on 12/Jun/20

$$\mathrm{Ok},\mathrm{each}\mathrm{time},\mathrm{i}\mathrm{will}\mathrm{think}\mathrm{how}\mathrm{to}\mathrm{transform}\mathrm{it}\mathrm{to}\mathrm{linear}.$$

Commented by Rio Michael last updated on 12/Jun/20

$$\mathrm{sir}\mathrm{please}\mathrm{any}\mathrm{hint}\mathrm{on}\mathrm{how}\mathrm{to}\mathrm{approach}Q98208??$$$$$$

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