JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
The sum of the coefficient of $x^{2/3}$ and $x^{-2/5}$ in the binomial expansion of $\left(x^{2/3}+\frac{1}{2}x^{-2/5}\right)^{9}$ is
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Find the power of $x$ in the general term, then the two values of $r$. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term. $$T_{r+1}={}^{9}C_r\,\frac{1}{2^r}\,x^{\frac23(9-r)-\frac25r}={}^{9}C_r\,\frac{1}{2^r}\,x^{6-\frac{16r}{15}}.$$ Step 2 - Coefficient of $x^{2/3}$. $6-\frac{16r}{15}=\frac23$ gives $r=5$: $\dfrac{{}^{9}C_5}{2^5}=\dfrac{126}{32}=\dfrac{63}{16}$. Step 3 - Coefficient of $x^{-2/5}$. $6-\frac{16r}{15}=-\frac25$ gives $r=6$: $\dfrac{{}^{9}C_6}{2^6}=\dfrac{84}{64}=\dfrac{21}{16}$. Step 4 - Sum. $\dfrac{63+21}{16}=\dfrac{84}{16}=\dfrac{21}{4}$. Answer: (A)