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If a_n=\frac{-2}{4 n^2-16 n+15},  then a_1+a_2+\cdots \ldots+a_{25} is equal to: 

 

Option: 1

\frac{52}{147}


Option: 2

\frac{49}{138}


Option: 3

\frac{50}{141}


Option: 4

\frac{51}{144}


Answers (1)

best_answer

given that a_n=\frac{-2}{4 n^2-16 n+15}

\begin{aligned} & a_1+a_2+a_3+\ldots . a_{25}=\sum_{n=1}^{25} \frac{-2}{(2 n-3)(2 n-5)} \\ & =\sum_{\mathrm{n}=1}^{25} \frac{(2 n-5)-(2 n-3)}{(2 n-3)(2 n-5)} \\ & =\sum_{\mathrm{n}=1}^{25}\left(\frac{1}{2 n-3}-\frac{1}{(2 n-5)}\right) \\ & =\frac{1}{-1}-\frac{1}{-3} \quad \end{aligned}

\begin{aligned} & +\frac{1}{1}-\frac{1}{-1} \\ & +\frac{1}{3}-\frac{1}{1} \\ & \vdots \quad \vdots \\ & \frac{1}{47}-\frac{1}{45} \\ & =\frac{1}{47}+\frac{1}{3} \\ & =\frac{3+47}{141}=\frac{50}{141} \end{aligned}

 

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Ritika Jonwal

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