JEE Main 2026 (Online) 5th April Morning ShiftQuestion 1
If the sum of the coefficients of $x^{7}$ and $x^{14}$ in the expansion of $\left(\frac{1}{x^{3}}-x^{4}\right)^{n}$, $x\neq0$, is zero, then the value of $n$ is ________.
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Answer$21$
Solution
Concept. The powers of $x$ go up in steps of $7$, so $x^7$ and $x^{14}$ come from consecutive terms. Formulas used. (1) General term of $(a+b)^n$: $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$(2) ${}^{n}C_r={}^{n}C_{r+1}\iff n=2r+1$Step 1 - General term. $$T_{r+1}={}^{n}C_r\,(x^{-3})^{n-r}(-x^4)^r=(-1)^r\,{}^{n}C_r\,x^{7r-3n}.$$ Step 2 - Which terms. $x^7$ needs $7r-3n=7$, i.e. $r=1+\frac{3n}{7}$. So $7\mid n$; write $n=7m$, then $r=3m+1$. The $x^{14}$ term is the next one, $r+1=3m+2$. Step 3 - Sum is zero. The two coefficients have opposite signs, so $$(-1)^r\left({}^{n}C_r-{}^{n}C_{r+1}\right)=0\ \Longrightarrow\ {}^{n}C_r={}^{n}C_{r+1}.$$ By (2), $n=2r+1$: $7m=2(3m+1)+1=6m+3$, so $m=3$. Step 4 - Value. $n=7m=21$. Answer: 21