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Binomial Theorem: JEE Main Mathematics Question with Solution

Q: Could there be other combinations of ?

No, because with forces , which leaves as the only possible integer.

The constant term in the expansion of

\left(2 x+\frac{1}{x^{7}}+3 x^{2}\right)^{5}

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Step-by-step solutionView

Correct answer

The constant term in the expansion is 1080.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Expression: $\left(2 x + \frac{1}{x^7} + 3 x^2\right)^5$ .

02goal

Find the constant term (the coefficient of $x^0$ ) in the expansion.

03approach

Use the multinomial theorem. The general term is $\frac{5!}{r_1! r_2! r_3!} (2x)^{r_1} (x^{-7})^{r_2} (3x^2)^{r_3}$ , with non-negative integers satisfying $r_1 + r_2 + r_3 = 5$ and power of $x$ equal to zero: $r_1 - 7r_2 + 2r_3 = 0$ .

04execute

From $r_1 + r_2 + r_3 = 5$ and $r_1 - 7r_2 + 2r_3 = 0$ , subtract the second from the first: $8r_2 - r_3 = 5 \implies r_3 = 8r_2 - 5$ . Since $r_1, r_2, r_3 \in \{0, 1, 2, 3, 4, 5\}$ , if $r_2 = 0$ , $r_3 = -5$ (invalid); if $r_2 = 1$ , $r_3 = 3$ , giving $r_1 = 5 - 1 - 3 = 1$ ; if $r_2 \ge 2$ , $r_3 \ge 11 > 5$ (invalid). Thus, the unique solution is $(r_1, r_2, r_3) = (1, 1, 3)$ .

05execute

Evaluate the coefficient for $(r_1, r_2, r_3) = (1, 1, 3)$ : $\text{Term} = \frac{5!}{1! \cdot 1! \cdot 3!} \cdot 2^1 \cdot 1^1 \cdot 3^3 = \frac{120}{6} \cdot 2 \cdot 27 = 20 \cdot 54 = 1080$ .

✓verify

Check the power of $x$ : $r_1 - 7r_2 + 2r_3 = 1 - 7(1) + 2(3) = 1 - 7 + 6 = 0$ . Sum of powers: $1 + 1 + 3 = 5$ . The solution is unique and fully verified.

Hints that build this answer step by step

Using the multinomial theorem, what is the general term in the expansion of $\left(2x + x^{-7} + 3x^2\right)^5$ ?
$\frac{5!}{a!b!c!} 2^a 3^c x^{a - 7b + 2c}$ where $a + b + c = 5$
Which non-negative integer values of $(a, b, c)$ satisfy $a + b + c = 5$ and make the power of $x$ zero ( $a - 7b + 2c = 0$ )?
$(a, b, c) = (1, 1, 3)$ only
What is the value of the coefficient for $(a, b, c) = (1, 1, 3)$ ?
1080

✓ 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

Could there be other combinations of

(r_1, r_2, r_3)

No, because $r_3 = 8r_2 - 5$ with $0 \le r_3 \le 5$ forces $5 \le 8r_2 \le 10$ , which leaves $r_2 = 1$ as the only possible integer.

Answer

The constant term in the expansion is 1080.

Why each option works or fails

Step-by-step solution

given: Expression: $\left(2 x + \frac{1}{x^7} + 3 x^2\right)^5$ .
goal: Find the constant term (the coefficient of $x^0$ ) in the expansion.
approach: Use the multinomial theorem. The general term is $\frac{5!}{r_1! r_2! r_3!} (2x)^{r_1} (x^{-7})^{r_2} (3x^2)^{r_3}$ , with non-negative integers satisfying $r_1 + r_2 + r_3 = 5$ and power of $x$ equal to zero: $r_1 - 7r_2 + 2r_3 = 0$ .
execute: From $r_1 + r_2 + r_3 = 5$ and $r_1 - 7r_2 + 2r_3 = 0$ , subtract the second from the first: $8r_2 - r_3 = 5 \implies r_3 = 8r_2 - 5$ . Since $r_1, r_2, r_3 \in \{0, 1, 2, 3, 4, 5\}$ , if $r_2 = 0$ , $r_3 = -5$ (invalid); if $r_2 = 1$ , $r_3 = 3$ , giving $r_1 = 5 - 1 - 3 = 1$ ; if $r_2 \ge 2$ , $r_3 \ge 11 > 5$ (invalid). Thus, the unique solution is $(r_1, r_2, r_3) = (1, 1, 3)$ .
execute: Evaluate the coefficient for $(r_1, r_2, r_3) = (1, 1, 3)$ : $\text{Term} = \frac{5!}{1! \cdot 1! \cdot 3!} \cdot 2^1 \cdot 1^1 \cdot 3^3 = \frac{120}{6} \cdot 2 \cdot 27 = 20 \cdot 54 = 1080$ .
verify: Check the power of $x$ : $r_1, r_2, r_3 \in \{0, 1, 2, 3, 4, 5\}$ . Sum of powers: $1 + 1 + 3 = 5$ . The solution is unique and fully verified.

Binomial Theorem: JEE Main Mathematics Question with Solution

What feels right?

The constant term in the expansion is 1080.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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