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Question Number 37915 by gunawan last updated on 19/Jun/18

$$\mathrm{If}y=4x^2-1,\mathrm{then}\mathrm{find}$$$$\frac{85}{169}+\underset{i=1}{\stackrel{84}{\mathrm{\Sigma }}}\frac{1}{y\left(i\right)}$$

Answered by ajfour last updated on 19/Jun/18

$$S-\frac{85}{169}=\underset{r=1}{\sum ^{84}}\frac{1}{(2r-1)(2r+1)}$$$$=\frac{1}{2}\underset{r=1}{\sum ^{84}}\left[\frac{(2r+1)-(2r-1)}{(2r-1)(2r+1)}\right]$$$$=\frac{1}{2}\underset{r=1}{\sum ^{84}}[\frac{1}{2r-1}-\frac{1}{2r+1}]$$$$=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{167}-\frac{1}{169})$$$$S=\frac{85}{169}-\frac{1}{2}(1-\frac{1}{169})=\frac{1}{169}.$$

Commented by gunawan last updated on 19/Jun/18

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}$$$$$$

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