Let the coefficients of x ^ - 1 and x ^ - 3 in the expansion of (2x(1)/(5) - (1)/(x(1)/(5))) ^ 15 , x > 0 , be m and n respectively. If r is a positive integer such m n ^ 2 = Cr . 15 2 ^ r , then the value of r is equal

Correct answer: 0 The r ^ th term of (2x(1)/(5) - (1)/(x(1)/(5))) ^ 15 ⇒ T _ r + 1 = Cr 15 (2x(1)/(5)) ^ 15 - r ((- 1)/(x(1)/(5))) ^ r For term having x ^ - 1 (15 - r)/(5) - (r)/(5) = - 1 ⇒ r = 10 ⇒ m = C10 15 · 2 ^ 5 · (- 1) ^ 10 For term having x ^ - 3 (15 - 2r)/(5) = - 3 ⇒ r = 15 ⇒ n = C15 15 · 2 ^ 0 · (- 1) ^ 15 ∴ m n ^ 2 = C10 15 · 2 ^ 5 = C5 15 · 2 ^ 5 On comparing with m · n ^ 2 = Cr 15 · 2 ^ r , we get r = 5

JEE Main 202229 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2022 · 29 June · Evening Shift · Mathematics · Binomial Theorem

Let the coefficients of x ^ - 1 and x ^ - 3 in the expansion of...

Let the coefficients of

{x}^{- 1}

and

{x}^{- 3}

in the expansion of

{\left(2x\dfrac{1}{5} - \dfrac{1}{x\dfrac{1}{5}}\right)}^{15}, x > 0

, be

m

and

n

respectively. If

r

is a positive integer such

m{n}^{2} = Cr . 15 {2}^{r}

, then the value of

r

is equal to ______.

Correct answer

Step-by-step solution

The ${r}^{th}$ term of ${\left(2x\dfrac{1}{5} - \dfrac{1}{x\dfrac{1}{5}}\right)}^{15}$

$⇒ {T}_{r + 1} = Cr 15 {\left(2x\dfrac{1}{5}\right)}^{15 - r}{\left(\dfrac{- 1}{x\dfrac{1}{5}}\right)}^{r}$

For term having ${x}^{- 1}$

$\dfrac{15 - r}{5} - \dfrac{r}{5} = - 1 ⇒ r = 10$ $⇒ m = C10 15 \cdot {2}^{5} \cdot {\left(- 1\right)}^{10}$

For term having ${x}^{- 3}$

$\dfrac{15 - 2r}{5} = - 3$ $⇒ r = 15$ $⇒ n = C15 15 \cdot {2}^{0} \cdot {\left(- 1\right)}^{15}$

$∴ m{n}^{2} = C10 15 \cdot {2}^{5} = C5 15 \cdot {2}^{5}$

On comparing with $m \cdot {n}^{2} = Cr 15 \cdot {2}^{r}$ , we get $r = 5$

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