If a _ n = - 2 4 n ^ 2 - 16n + 15 , then a _ 1 + a _ 2 + . + a _ 25 is equal to:

Correct answer: C. (50)/(141) We have, a _ n = - 2 4 n ^ 2 - 16n + 15 ⇒ a _ n = - (2)/((2n - 5)(2n - 3)) ⇒ a _ n = - (1)/(2)[(1)/((n - (5)/(2))(n - (3)/(2)))] ⇒ a _ n = - (1)/(2)[(1)/((n - (5)/(2))) - (1)/((n - (3)/(2)))] ⇒ a _ n = (1)/(2)[(1)/((n - (3)/(2))) - (1)/((n - (5)/(2)))] ⇒ ∑ _ n = 1 ^ 25 a _ n = (1)/(2) ∑ _ n = 1 ^ 25 [(1)/((n - (3)/(2))) - (1)/((n - (5)/(2)))] ⇒ ∑ _ n = 1 ^ 25 a _ n = (1)/(2)[\ (1)/((- (1)/(2))) - (1)/((- (3)/(2)))\ + \ (1)/(((1)/(2))) - (1)/((- (1)/(2)))\ + \ (1)/(((3)/(2))) - (1)/(((1)/(2)))\ + \ (1)/(((5)/(2))) - (1)/(((3)/(2)))\ + \ (1)/(((7)/(2))) - (1)/(((5)/(2)))\ + . . . + \ (1)/(((45)/(2))) - (1)/(((43)/(2)))\ + \ (1)/(((47)/(2))) - (1)/(((45)/(2)))\ ] ⇒ ∑ _ n = 1 ^ 25 a _ n = (1)/(2)((2)/(3) + (2)/(47)) ⇒ ∑ _ n = 1 ^ 25 a _ n = ((1)/(3) + (1)/(47)) ⇒ ∑ _ n = 1 ^ 25 a _ n = (50)/(141)

JEE Main 202330 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 30 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2023 · 30 January · Morning Shift · Mathematics · Sequences and Series

If a _ n = - 2 4 n ^ 2 - 16n + 15 , then a _ 1 + a _ 2 + . + a _ 25...

{a}_{n} = \dfrac{- 2}{4{n}^{2} - 16n + 15}

, then

{a}_{1} + {a}_{2} + \ldots . + {a}_{25}

is equal to:

Options

A $\dfrac{51}{144}$
B $\dfrac{49}{138}$
C $\dfrac{50}{141}$
D $\dfrac{52}{147}$

Correct answer

C. $\dfrac{50}{141}$

Step-by-step solution

We have,

${a}_{n} = - \dfrac{2}{4{n}^{2} - 16n + 15}$

$⇒ {a}_{n} = - \dfrac{2}{\left(2n - 5\right)\left(2n - 3\right)}$

$⇒ {a}_{n} = - \dfrac{1}{2}\left[\dfrac{1}{\left(n - \dfrac{5}{2}\right)\left(n - \dfrac{3}{2}\right)}\right]$

$⇒ {a}_{n} = - \dfrac{1}{2}\left[\dfrac{1}{\left(n - \dfrac{5}{2}\right)} - \dfrac{1}{\left(n - \dfrac{3}{2}\right)}\right]$

$⇒ {a}_{n} = \dfrac{1}{2}\left[\dfrac{1}{\left(n - \dfrac{3}{2}\right)} - \dfrac{1}{\left(n - \dfrac{5}{2}\right)}\right]$

$⇒ \mathop{∑}_{n = 1}^{25}{a}_{n} = \dfrac{1}{2}\mathop{∑}_{n = 1}^{25}\left[\dfrac{1}{\left(n - \dfrac{3}{2}\right)} - \dfrac{1}{\left(n - \dfrac{5}{2}\right)}\right]$

$⇒ \mathop{∑}_{n = 1}^{25}{a}_{n} = \dfrac{1}{2}\left[\left\{\dfrac{1}{\left(- \dfrac{1}{2}\right)} - \dfrac{1}{\left(- \dfrac{3}{2}\right)}\right\} + \left\{\dfrac{1}{\left(\dfrac{1}{2}\right)} - \dfrac{1}{\left(- \dfrac{1}{2}\right)}\right\} + \left\{\dfrac{1}{\left(\dfrac{3}{2}\right)} - \dfrac{1}{\left(\dfrac{1}{2}\right)}\right\} + \left\{\dfrac{1}{\left(\dfrac{5}{2}\right)} - \dfrac{1}{\left(\dfrac{3}{2}\right)}\right\} + \left\{\dfrac{1}{\left(\dfrac{7}{2}\right)} - \dfrac{1}{\left(\dfrac{5}{2}\right)}\right\} + . . . + \left\{\dfrac{1}{\left(\dfrac{45}{2}\right)} - \dfrac{1}{\left(\dfrac{43}{2}\right)}\right\} + \left\{\dfrac{1}{\left(\dfrac{47}{2}\right)} - \dfrac{1}{\left(\dfrac{45}{2}\right)}\right\}\right]$

$⇒ \mathop{∑}_{n = 1}^{25}{a}_{n} = \dfrac{1}{2}\left(\dfrac{2}{3} + \dfrac{2}{47}\right)$

$⇒ \mathop{∑}_{n = 1}^{25}{a}_{n} = \left(\dfrac{1}{3} + \dfrac{1}{47}\right)$

$⇒ \mathop{∑}_{n = 1}^{25}{a}_{n} = \dfrac{50}{141}$

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