Binomial TheoremmediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Binomial Theorem: Coefficient Expansion (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
The coefficient of x2x^2 in the expansion of (2x2+1x)10\left(2x^2+\dfrac{1}{x}\right)^{10}, x≠0x\neq0, is:
A32403240
B33603360correct
C34803480
D36003600
Solution
Tr+1=(10r)(2x2)10−r ⁣(1x)r=(10r)210−r x20−3rT_{r+1} = \binom{10}{r}(2x^2)^{10-r}\!\left(\frac{1}{x}\right)^r = \binom{10}{r}2^{10-r}\,x^{20-3r}
 Step 2: Determine rr for the x2x^2 term
20−3r=2  ⟹  r=620-3r = 2 \implies r = 6
 Step 3: Compute the coefficient
(106)⋅24=210×16=3360\binom{10}{6}\cdot2^{4} = 210\times16 = 3360
 Answer: (2)
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