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Question Number 102348 by mohammad17 last updated on 08/Jul/20

Answered by mr W last updated on 08/Jul/20

$$a_n=\frac{{(2n+1)}^2+1}{{(2n+1)}^2-1}$$$$=1+\frac{2}{{(2n+1)}^2-1}$$$$=1+\frac{2}{\left(2n\right)(2n+2)}$$$$=1+\frac{1}{2}\times \frac{1}{n(n+1)}$$$$=1+\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})$$$$\underset{n=1}{\sum ^{49}}{a}_n=49+\frac{1}{2}(\frac{1}{1}-\frac{1}{50})=49+\frac{49}{100}=49.49$$

Commented by mohammad17 last updated on 08/Jul/20

$$nicesirthankyou$$

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