JEE Main 2026 (Online) 5th April Evening ShiftQuestion 1
The coefficient of $x^{2}$ in the expansion of $\left(2x^{2}+\frac{1}{x}\right)^{10}$, $x\neq0$, is :
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Write the general term and choose $r$ so that the power of $x$ is $2$. Formulas used. (1) General term of $(a+b)^n$: $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term. $$T_{r+1}={}^{10}C_r\,(2x^2)^{10-r}\left(\frac1x\right)^r={}^{10}C_r\,2^{10-r}\,x^{20-3r}.$$ Step 2 - Power $2$. $20-3r=2$ gives $r=6$. Step 3 - Coefficient. $${}^{10}C_6\cdot2^{4}=210\cdot16=3360.$$ Answer: (B)