JEE Main 2026 Apr 5 Shift 2, Mathematics Q7: Introduction to Binomial Expansion

JEE Main2026Apr 5, Shift 2Mathematics

The coefficient of $x^{2}$ in the expansion of $(2 x^{2} + \frac{1}{x})^{10}$ , $x \neq = 0$ , is:

Solution

The general term: $T_{r + 1} = (r 10) (2 x^{2})^{10 - r} (\frac{1}{x})^{r} = (r 10) 2^{10 - r} x^{20 - 3 r}$ .

For the $x^{2}$ term: $20 - 3 r = 2 \Rightarrow r = 6$ .

Coefficient $= (6 10) \cdot 2^{4} = 210 \cdot 16 = 3360$ .

Concept behind this question

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