If ∑ _ r = 1 ^ n t _ r = (n(n + 1)(n + 2)(n + 3))/(8), where t _ r denotes the r ^ th term of a series, then _ n → ∞ ∑ _ r = 1 ^ n 1 t _ r is

Correct answer: C. (1)/(2) Given, ∑ _ r = 1 ^ n t _ r = (n(n + 1)(n + 2)(n + 3))/(8) = S _ n (say) ∴ ∑ _ r = 1 ^ n - 1 t _ r = ((n - 1)n(n + 1)(n + 2))/(8) = S _ n - 1 Now, t _ n = S _ n - S _ n - 1 = (n(n + 1)(n + 2))/(2) ∴ _ n → ∞ ∑ _ r = 1 ^ n 1 t _ r = _ n → ∞ ∑ _ r = 1 ^ n (2)/(n(n + 1)(n + 2)) = _ n → ∞ ∑ _ r = 1 ^ n ((1)/(n(n + 1)) - (1)/((n + 1)(n + 2))) = - _ n → ∞ ∑ _ r = 1 ^ n ((1)/((n + 1)(n + 2)) - (1)/(n(n + 1))) = - _ n → ∞ ((1)/((n + 1)(n + 2)) - (1)/(2)) = - (0 - (1)/(2)) = (1)/(2)

BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Sequences and Series

BITSAT 2018 · Mathematics · Sequences and Series

If ∑ _ r = 1 ^ n t _ r = (n(n + 1)(n + 2)(n + 3))/(8), where t _ r...

\mathop{∑}_{r = 1}^{n}{t}_{r} = \dfrac{n\left(n + 1\right)\left(n + 2\right)\left(n + 3\right)}{8},

where

{t}_{r}

denotes the

{r}^{th}

term of a series, then

\lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{1}{{t}_{r}}

Options

A $\dfrac{1}{8}$
B $\dfrac{1}{4}$
C $\dfrac{1}{2}$
D $1$

Correct answer

C. $\dfrac{1}{2}$

Step-by-step solution

Given, $\mathop{∑}_{r = 1}^{n}{t}_{r} = \dfrac{n\left(n + 1\right)\left(n + 2\right)\left(n + 3\right)}{8} = {S}_{n}$ (say)

$∴ \mathop{∑}_{r = 1}^{n - 1}{t}_{r} = \dfrac{\left(n - 1\right)n\left(n + 1\right)\left(n + 2\right)}{8} = {S}_{n - 1}$

Now, ${t}_{n} = {S}_{n} - {S}_{n - 1} = \dfrac{n\left(n + 1\right)\left(n + 2\right)}{2}$

$∴ \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{1}{{t}_{r}} = \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{2}{n\left(n + 1\right)\left(n + 2\right)}$

$= \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\left(\dfrac{1}{n\left(n + 1\right)} - \dfrac{1}{\left(n + 1\right)\left(n + 2\right)}\right)$

$= - \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\left(\dfrac{1}{\left(n + 1\right)\left(n + 2\right)} - \dfrac{1}{n\left(n + 1\right)}\right)$

$= - \lim_{n \to \infty}\left(\dfrac{1}{\left(n + 1\right)\left(n + 2\right)} - \dfrac{1}{2}\right)$

$= - \left(0 - \dfrac{1}{2}\right) = \dfrac{1}{2}$

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