JEE MainMathematics

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Finding a Specific Term in a Binomial Expansion

Q: Why is the power of x written as 9 - 3r?

Because the first term contributes x^(9-r) and the second term has x^2 in the denominator, contributing (x^(-2))^r = x^(-2r). Multiplying them gives x^((9-r) - 2r) = x^(9-3r).

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

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Correct answer

The coefficient of $x^{- 6}$ in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ is $5040$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The expression is ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , and we need the coefficient of $x^{- 6}$ .

02Goal

Find the numerical value of the coefficient of $x^{- 6}$ .

03Approach

Write down the general term $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ . Collect powers of $x$ to form an equation for the exponent equal to $- 6$ . Solve for $r$ , then evaluate the constant factor of $T_{r + 1}$ .

04Execute

The general term is: $T_{r + 1} = (\binom{9}{r}) {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r} = (\binom{9}{r}) {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} x^{9 - r - 2 r} = (\binom{9}{r}) {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} x^{9 - 3 r}$ Set the exponent of $x$ equal to $- 6$ : $9 - 3 r = - 6 ⟹ 3 r = 15 ⟹ r = 5$

05Execute

Substitute $r = 5$ into the coefficient expression: $Coefficient = (\binom{9}{5}) {(\frac{4}{5})}^{9 - 5} {(\frac{5}{2})}^{5} = (\binom{9}{4}) {(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5}$ Evaluate the terms: $(\binom{9}{4}) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ ${(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5} = \frac{4^{4} \times 5^{5}}{5^{4} \times 2^{5}} = \frac{2^{8} \times 5}{2^{5}} = 2^{3} \times 5 = 8 \times 5 = 40$ $Coefficient = 126 \times 40 = 5040$

✓Verify

Double check: $(\binom{9}{5}) = 126$ . Powers of 5: $5^{5} / 5^{4} = 5$ . Powers of 2: $4^{4} / 2^{5} = 2^{8} / 2^{5} = 2^{3} = 8$ . Product is $126 \times 40 = 5040$ . Value is an exact integer as expected.

Hints that build this answer step by step

Using the binomial theorem, what is the general term $T_{r + 1}$ in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ ?
$T_{r + 1} = (\binom{9}{r}) {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r}$
What value of $r$ corresponds to the term containing $x^{- 6}$ ?
$r = 5$
What is the numerical coefficient when $r = 5$ is evaluated in $(\binom{9}{5}) {(\frac{4}{5})}^{9 - 5} {(\frac{5}{2})}^{5}$ ?
$5040$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why is the power of

x

written as

9 - 3 r

Because the first term contributes $x^{9 - r}$ and the second term has $x^{2}$ in the denominator, contributing $(x^{- 2})^{r} = x^{- 2 r}$ . Multiplying them gives $x^{(9 - r) - 2 r} = x^{9 - 3 r}$ .

Answer

The coefficient of $x^{- 6}$ in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ is $5040$ .

Why each option works or fails

Step-by-step solution

Given: The expression is ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ , and we need the coefficient of $x^{- 6}$ .
Goal: Find the numerical value of the coefficient of $x^{- 6}$ .
Approach: Write down the general term $T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$ . Collect powers of $x$ to form an equation for the exponent equal to $- 6$ . Solve for $r$ , then evaluate the constant factor of $T_{r + 1}$ .
Execute: The general term is: $T_{r + 1} = (\binom{9}{r}) {(\frac{4 x}{5})}^{9 - r} {(\frac{5}{2 x^{2}})}^{r} = (\binom{9}{r}) {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} x^{9 - r - 2 r} = (\binom{9}{r}) {(\frac{4}{5})}^{9 - r} {(\frac{5}{2})}^{r} x^{9 - 3 r}$ Set the exponent of $x$ equal to $- 6$ : $9 - 3 r = - 6 ⟹ 3 r = 15 ⟹ r = 5$
Execute: Substitute $r = 5$ into the coefficient expression: $Coefficient = (\binom{9}{5}) {(\frac{4}{5})}^{9 - 5} {(\frac{5}{2})}^{5} = (\binom{9}{4}) {(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5}$ Evaluate the terms: $(\binom{9}{4}) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ ${(\frac{4}{5})}^{4} {(\frac{5}{2})}^{5} = \frac{4^{4} \times 5^{5}}{5^{4} \times 2^{5}} = \frac{2^{8} \times 5}{2^{5}} = 2^{3} \times 5 = 8 \times 5 = 40$ $Coefficient = 126 \times 40 = 5040$
Verify: Double check: $(\binom{9}{5}) = 126$ . Powers of 5: $5^{5} / 5^{4} = 5$ . Powers of 2: $4^{4} / 2^{5} = 2^{8} / 2^{5} = 2^{3} = 8$ . Product is $126 \times 40 = 5040$ . Value is an exact integer as expected.

Finding a Specific Term in a Binomial Expansion

What feels right?

The coefficient of $x^{- 6}$ in the expansion of ${(\frac{4 x}{5} + \frac{5}{2 x^{2}})}^{9}$ is $5040$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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