If a n = - 2 4 n 2 - 16 n + 15 , then a 1 + a 2 + . . . + a 25 is equal to [2023]

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Solution

Q.
If $a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15}$ , then $a_{1} + a_{2} + . . . + a_{25}$ is equal to [2023]

1 52/147

2 50/141

3 51/144

4 49/138

Ans.
(2)

$a_{n} = \frac{- 2}{4 n^{2} - 16 n + 15} = \frac{- 2}{(2 n - 3) (2 n - 5)}$

$a_{n} = \frac{1}{2 n - 3} - \frac{1}{2 n - 5}$

$For n = 1, a_{1} = - 1 + \frac{1}{3}; n = 2, a_{2} = 1 + 1; n = 3, a_{3} = \frac{1}{3} - 1$

$n = 4, a_{4} = \frac{1}{5} - \frac{1}{3}; n = 5, a_{5} = \frac{1}{7} - \frac{1}{5}$

$n = 25, a_{25} = \frac{1}{47} - \frac{1}{45}$

$Now, a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + \dots + a_{25}$

$= (- 1 + \frac{1}{3}) + (1 + 1) + (\frac{1}{3} - 1) + (\frac{1}{5} - \frac{1}{3}) + (\frac{1}{7} - \frac{1}{5}) + \dots + (\frac{1}{47} - \frac{1}{45})$

$= - 1 + \frac{1}{3} + 2 - 1 + \frac{1}{47} = \frac{50}{141}$

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