Question Number 134269 by I want to learn more last updated on 01/Mar/21
$$\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\:\:\:\frac{\mathrm{k}^{\mathrm{2}} \:\:+\:\:\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \:\:−\:\:\mathrm{1}} \\ $$
Answered by mr W last updated on 01/Mar/21
$$=\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\:\:\:\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{1}\:\:+\:\:\mathrm{2}}{\mathrm{k}^{\mathrm{2}} \:\:−\:\:\mathrm{1}} \\ $$$$=\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\left(\mathrm{1}+\:\:\:\frac{\:\:\mathrm{2}}{\mathrm{k}^{\mathrm{2}} \:\:−\:\:\mathrm{1}}\right) \\ $$$$=\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\left(\mathrm{1}+\:\:\:\frac{\:\:\mathrm{1}}{\mathrm{k}\:\:−\:\:\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right) \\ $$$$=\mathrm{98}+\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{\mathrm{98}}\right)−\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{98}}+\frac{\mathrm{1}}{\mathrm{99}}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$=\mathrm{98}+\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}}{\mathrm{99}}+\frac{\mathrm{1}}{\mathrm{100}}\right) \\ $$$$=\mathrm{99}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{100}} \\ $$$$=\mathrm{99}\frac{\mathrm{4751}}{\mathrm{9900}} \\ $$
Commented by I want to learn more last updated on 02/Mar/21
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