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

Σ_(k = 2) ^(99) ((k^2 + 1)/(k^2 − 1))

$$\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\:\:\:\frac{\mathrm{k}^{\mathrm{2}} \:\:+\:\:\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \:\:−\:\:\mathrm{1}} \\ $$

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

=Σ_(k = 2) ^(99) ((k^2 −1 + 2)/(k^2 − 1)) =Σ_(k = 2) ^(99) (1+ (( 2)/(k^2 − 1))) =Σ_(k = 2) ^(99) (1+ (( 1)/(k − 1))−(1/(k+1))) =98+((1/1)+(1/2)+(1/3)+...+(1/(98)))−((1/3)+(1/4)+...+(1/(98))+(1/(99))+(1/(100))) =98+((1/1)+(1/2))−((1/(99))+(1/(100))) =99+(1/2)−(1/(99))−(1/(100)) =99((4751)/(9900))

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

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

Thanks sir. i really appreciate. God bless you sir.

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

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