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Question Number 134269 by I want to learn more last updated on 01/Mar/21

$$\underset{\mathrm{k}=2}{\sum ^{99}}\frac{{\mathrm{k}}^2+1}{{\mathrm{k}}^2-1}$$

Answered by mr W last updated on 01/Mar/21

$$=\underset{\mathrm{k}=2}{\sum ^{99}}\frac{{\mathrm{k}}^2-1+2}{{\mathrm{k}}^2-1}$$$$=\underset{\mathrm{k}=2}{\sum ^{99}}(1+\frac{2}{{\mathrm{k}}^2-1})$$$$=\underset{\mathrm{k}=2}{\sum ^{99}}(1+\frac{1}{\mathrm{k}-1}-\frac{1}{k+1})$$$$=98+(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98})-(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}+\frac{1}{100})$$$$=98+(\frac{1}{1}+\frac{1}{2})-(\frac{1}{99}+\frac{1}{100})$$$$=99+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}$$$$=99\frac{4751}{9900}$$

Commented by I want to learn more last updated on 02/Mar/21

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{i}\mathrm{really}\mathrm{appreciate}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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