JEE Main 2021 (Online) 31st August Morning ShiftQuestion 1
If $\left(\frac{3^{6}}{4^{4}}\right)k$ is the term, independent of $x$, in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$, then $k$ is equal to $\_\_\_\_\_\_$ .
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$55$
Solution
Concept. Find $r$ that removes $x$, then simplify the powers of $2$ and $3$. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$Step 1 - General term. $$T_{r+1}={}^{12}C_r\left(\frac x4\right)^{12-r}\left(-\frac{12}{x^2}\right)^r,\qquad\text{power of }x:\ 12-3r.$$ Step 2 - Power zero. $r=4$: $${}^{12}C_4\cdot\frac{12^4}{4^8}=495\cdot\frac{2^8\cdot3^4}{2^{16}}=\frac{495\cdot81}{256}.$$ Step 3 - Match $\frac{3^6}{4^4}k$. $4^4=256$, so $729k=495\cdot81$ and $$k=\frac{495\cdot81}{729}=\frac{495}{9}=55.$$ Answer: 55