Why is the power of written as ?

Because the first term contributes and the second term has in the denominator, contributing . Multiplying them gives .

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Binomial Theorem: JEE Main Mathematics Question with Solution

The coefficient of

x^{-6}

, in the expansion of

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

, is

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Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

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

05execute

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

✓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 $\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$ ?
$T_{r+1} = \binom{9}{r} \left(\frac{4x}{5}\right)^{9-r} \left(\frac{5}{2x^2}\right)^r$
What value of $r$ corresponds to the term containing $x^{-6}$ ?
$r = 5$
What is the numerical coefficient when $r = 5$ is evaluated in $\binom{9}{5} \left(\frac{4}{5}\right)^{9-5} \left(\frac{5}{2}\right)^5$ ?
$5040$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

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 $x^{(9-r) - 2r} = x^{9-3r}$ .

Answer

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

Why each option works or fails

Step-by-step solution

given: The expression is $\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$ , 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: $T_{r+1} = \binom{9}{r} \left(\frac{4x}{5}\right)^{9-r} \left(\frac{5}{2x^2}\right)^r = \binom{9}{r} \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^r x^{9-r-2r} = \binom{9}{r} \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^r x^{9-3r}$ Set the exponent of $x$ equal to $-6$ : $9 - 3r = -6 \implies 3r = 15 \implies r = 5$
execute: Substitute $r = 5$ into the coefficient expression: $\text{Coefficient} = \binom{9}{5} \left(\frac{4}{5}\right)^{9-5} \left(\frac{5}{2}\right)^5 = \binom{9}{4} \left(\frac{4}{5}\right)^4 \left(\frac{5}{2}\right)^5$ Evaluate the terms: $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ $\left(\frac{4}{5}\right)^4 \left(\frac{5}{2}\right)^5 = \frac{4^4 \times 5^5}{5^4 \times 2^5} = \frac{2^8 \times 5}{2^5} = 2^3 \times 5 = 8 \times 5 = 40$ $\text{Coefficient} = 126 \times 40 = 5040$
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.

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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