If a_n = frac{-2}{4n^2 - 16n + 15},then a_1 + | vedclass

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(C) Given $a_n = \frac{-2}{4n^2 - 16n + 15}$.Factorizing the denominator: $4n^2 - 16n + 15 = 4n^2 - 6n - 10n + 15 = 2n(2n - 3) - 5(2n - 3) = (2n - 3)(

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$\frac{50}{141}$

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(C) Given $a_n = \frac{-2}{4n^2 - 16n + 15}$.Factorizing the denominator: $4n^2 - 16n + 15 = 4n^2 - 6n - 10n + 15 = 2n(2n - 3) - 5(2n - 3) = (2n - 3)(2n - 5)$.Thus,$a_n = \frac{-2}{(2n - 3)(2n - 5)} = \frac{1}{2n - 5} - \frac{1}{2n - 3}$.This is a telescoping series.Sum $S_{25} = \sum_{n=1}^{25} \left( \frac{1}{2n - 5} - \frac{1}{2n - 3} \right)$.$S_{25} = \left( \frac{1}{-3} - \frac{1}{-1} \right) + \left( \frac{1}{-1} - \frac{1}{1} \right) + \dots + \left( \frac{1}{45} - \frac{1}{47} \right)$.$S_{25} = \frac{1}{-3} - \frac{1}{47} = -\frac{1}{3} - \frac{1}{47} = \frac{-47 - 3}{141} = -\frac{50}{141}$.Wait,checking the partial fraction decomposition: $\frac{1}{2n-5} - \frac{1}{2n-3} = \frac{(2n-3) - (2n-5)}{(2n-5)(2n-3)} = \frac{2}{(2n-5)(2n-3)}$.Since $a_n = \frac{-2}{(2n-3)(2n-5)}$,we have $a_n = \frac{1}{2n-3} - \frac{1}{2n-5}$.Sum $S_{25} = \sum_{n=1}^{25} \left( \frac{1}{2n-3} - \frac{1}{2n-5} \right) = \left( \frac{1}{-1} - \frac{1}{-3} \right) + \left( \frac{1}{1} - \frac{1}{-1} \right) + \dots + \left( \frac{1}{47} - \frac{1}{45} \right) = \frac{1}{47} - \frac{1}{-3} = \frac{1}{47} + \frac{1}{3} = \frac{3 + 47}{141} = \frac{50}{141}$.

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

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