The coefficient of x ^ - 6 , in the expansion of ((4x)/(5) + 5 2 x ^ 2 ) ^ 9 , is

Correct answer: 5040 Given, ((4x)/(5) + 5 2 x ^ 2 ) ^ 9 Now r ^ th term of the expression is given by, T _ r + 1 = Cr 9 ((4x)/(5)) ^ 9 - r ( 5 2 x ^ 2 ) ^ r ⇒ T _ r + 1 = Cr 9 ((4)/(5)) ^ 9 - r ((5)/(2)) ^ r x ^ 9 - r - 2r Now equating 9 - r - 2r = - 6 ⇒ r = 5 So, coefficient of x ^ - 6 will be, = C5 9 ((4)/(5)) ^ 9 - 5 ((5)/(2)) ^ 5 = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) × ((4)/(5)) ^ 4 ((5)/(2)) ^ 5 = 5040

JEE Main 202331 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 31 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2023 · 31 January · Evening Shift · Mathematics · Binomial Theorem

The coefficient of x ^ - 6 , in the expansion of ((4x)/(5) + 5 2 x...

The coefficient of

{x}^{- 6}

, in the expansion of

{\left(\dfrac{4x}{5} + \dfrac{5}{2{x}^{2}}\right)}^{9}

, is

Correct answer

5040

Step-by-step solution

Given,

${\left(\dfrac{4x}{5} + \dfrac{5}{2{x}^{2}}\right)}^{9}$

Now ${r}^{th}$ term of the expression is given by,

${T}_{r + 1} = Cr 9 {\left(\dfrac{4x}{5}\right)}^{9 - r}{\left(\dfrac{5}{2{x}^{2}}\right)}^{r}$

$⇒ {T}_{r + 1} = Cr 9 {\left(\dfrac{4}{5}\right)}^{9 - r}{\left(\dfrac{5}{2}\right)}^{r}{x}^{9 - r - 2r}$

Now equating $9 - r - 2r = - 6 ⇒ r = 5$

So, coefficient of ${x}^{- 6}$ will be,

$= C5 9 {\left(\dfrac{4}{5}\right)}^{9 - 5}{\left(\dfrac{5}{2}\right)}^{5}$

$= \dfrac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times {\left(\dfrac{4}{5}\right)}^{4}{\left(\dfrac{5}{2}\right)}^{5}$

$= 5040$

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