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

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

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

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

= \underset{k = 2}{\sum^{99}} (1 + \frac{2}{k^{2} - 1})

= \underset{k = 2}{\sum^{99}} (1 + \frac{1}{k - 1} - \frac{1}{k + 1})

= 98 + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{98}) - (\frac{1}{3} + \frac{1}{4} + \dots + \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

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