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Question Number 194015 by Rupesh123 last updated on 26/Jun/23

Answered by witcher3 last updated on 26/Jun/23

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}⩾1}{\mathrm{f}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}$$$$f_{k+1}=f_k+f_{k-1};{\mathrm{f}}_1={\mathrm{f}}_2=1$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+{\mathrm{x}}^2+\sum _{\mathrm{k}⩾2}{\mathrm{f}}_{\mathrm{k}+1}{\mathrm{x}}^{\mathrm{k}+1}$$$$\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}-{\mathrm{x}}^2=\mathrm{x}\sum _{\mathrm{k}⩾2}({\mathrm{f}}_{\mathrm{k}}+{\mathrm{f}}_{\mathrm{k}-1}){\mathrm{x}}^{\mathrm{k}}$$$$\iff \mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}-{\mathrm{x}}^2=\mathrm{x}\sum _{\mathrm{k}⩾2}{\mathrm{f}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}+{\mathrm{x}}^2\sum _{\mathrm{k}⩾2}{\mathrm{f}}_{\mathrm{k}-1}{\mathrm{x}}^{\mathrm{k}-1}$$$$\iff \mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}-{\mathrm{x}}^2=\mathrm{x}(\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x})+{\mathrm{x}}^2\mathrm{f}\left(\mathrm{x}\right)$$$$\iff \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{1-\mathrm{x}-{\mathrm{x}}^2}$$$$\mathrm{f}\left(\frac{1}{10}\right)=\sum _{\mathrm{k}⩾1}\frac{{\mathrm{x}}^{\mathrm{k}}}{{10}^{\mathrm{k}}}{\mathrm{f}}_{\mathrm{k}}\Rightarrow \frac{1}{10}\mathrm{f}\left(\frac{1}{10}\right)=\mathrm{S}=\frac{\frac{1}{10}}{1-\frac{1}{10}-\frac{1}{100}}.\frac{1}{10}$$$$=\frac{1}{89}$$

Commented by Rupesh123 last updated on 26/Jun/23

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