JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
The absolute difference of the coefficients of $x^{10}$ and $x^{7}$ in the expansion of $\left(2x^{2}+\frac{1}{2x}\right)^{11}$ is equal to
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Find both coefficients from the general term and subtract. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term. $$T_{r+1}={}^{11}C_r(2x^2)^{11-r}\left(\frac{1}{2x}\right)^r={}^{11}C_r\,2^{11-2r}\,x^{22-3r}.$$ Step 2 - $x^{10}$. $r=4$: ${}^{11}C_4\cdot2^3=330\cdot8=2640$. Step 3 - $x^{7}$. $r=5$: ${}^{11}C_5\cdot2=462\cdot2=924$. Step 4 - Difference. $2640-924=1716=1728-12=12^3-12$. Answer: (B)