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Question Number 129009 by bramlexs22 last updated on 12/Jan/21

$$\mathrm{Given}\mathrm{f}(\mathrm{x}+1)=\frac{1+\mathrm{f}\left(\mathrm{x}\right)}{1-\mathrm{f}\left(\mathrm{x}\right)};\mathrm{f}\left(2\right)=2$$$$\mathrm{and}{\int }_2^{2018}\mathrm{x}.\mathrm{f}\left(2018\right)\mathrm{dx}=2^{\mathrm{a}}.3^{\mathrm{b}}.5^{\mathrm{c}}.7^{\mathrm{d}}.{11}^{\mathrm{e}}.{101}^{\mathrm{f}}$$$$\mathrm{then}\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=\mathrm{\_}\mathrm{\_}$$

Answered by liberty last updated on 12/Jan/21

$$\to \{\begin{array}{l}\mathrm{x}=1\to \mathrm{f}\left(2\right)=\frac{1+\mathrm{f}\left(1\right)}{1-\mathrm{f}\left(1\right)}=2;\mathrm{f}\left(1\right)=\frac{1}{3}\\ \mathrm{x}=2\Rightarrow \mathrm{f}\left(3\right)=\frac{1+\mathrm{f}\left(2\right)}{1-\mathrm{f}\left(2\right)}=\frac{3}{-1}=-3\\ \mathrm{x}=3\Rightarrow \mathrm{f}\left(4\right)=\frac{1+\mathrm{f}\left(3\right)}{1-\mathrm{f}\left(3\right)}=\frac{-2}{4}=-\frac{1}{2}\\ \mathrm{x}=4\Rightarrow \mathrm{f}\left(5\right)=\frac{1+\mathrm{f}\left(4\right)}{1-\mathrm{f}\left(4\right)}=\frac{1}{3}\\ \mathrm{x}=5\Rightarrow \mathrm{f}\left(6\right)=\frac{1+\mathrm{f}\left(5\right)}{1-\mathrm{f}\left(5\right)}=\frac{4/3}{2/3}=2\end{array}$$$$\Rightarrow \frac{1}{3};2;-3;-\frac{1}{2};\frac{1}{3};2;...$$$$\mathrm{then}\mathrm{f}\left(2018\right)=\mathrm{f}(4\times 504+2)=\mathrm{f}\left(2\right)=2$$$${\int }_2^{2018}\mathrm{xf}\left(2018\right)\mathrm{dx}={\int }_2^{2018}2\mathrm{x}\mathrm{dx}={2018}^2-2^2$$$$=2020\times 2016=4\times 505\times 4\times 504$$$$=2^4\times 5\times 101\times 3\times 168$$$$=2^4\times 5\times 101\times 7\times 2^3\times 3^2$$$$=2^7\times 3^2\times 5\times 7\times 101$$

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