JEE Main 2026 — Mathematics Algebra

The general term in the expansion of $\left(\dfrac{1}{x^3} - x^4\right)^n$(x31​−x4)n is given by:

$$T_{r+1} = ^{n}C_{r} \left(x^{-3}\right)^{n-r} \left(-x^4\right)^r = (-1)^r \cdot ^{n}C_{r} \cdot x^{7r - 3n}$$Tr+1​=nCr​(x−3)n−r(−x4)r=(−1)r⋅nCr​⋅x7r−3n

For the coefficient of $x^7$x7, we set the exponent to $7$7:

$$7r_1 - 3n = 7 \Rightarrow r_1 = \dfrac{3n + 7}{7}$$7r1​−3n=7⇒r1​=73n+7​

For the coefficient of $x^{14}$x14, we set the exponent to $14$14:

$$7r_2 - 3n = 14 \Rightarrow r_2 = \dfrac{3n + 14}{7}$$7r2​−3n=14⇒r2​=73n+14​

Notice that $r_2 = r_1 + 1$r2​=r1​+1.

The sum of the coefficients of $x^7$x7 and $x^{14}$x14 is zero:

$$(-1)^{r_1} \cdot ^{n}C_{r_1} + (-1)^{r_2} \cdot ^{n}C_{r_2} = 0$$(−1)r1​⋅nCr1​​+(−1)r2​⋅nCr2​​=0

$$(-1)^{r_1} \cdot ^{n}C_{r_1} + (-1)^{r_1 + 1} \cdot ^{n}C_{r_1 + 1} = 0$$(−1)r1​⋅nCr1​​+(−1)r1​+1⋅nCr1​+1​=0

$$(-1)^{r_1} \left( ^{n}C_{r_1} - ^{n}C_{r_1 + 1} \right) = 0$$(−1)r1​(nCr1​​−nCr1​+1​)=0

$$^{n}C_{r_1} = ^{n}C_{r_1 + 1}$$nCr1​​=nCr1​+1​

Since $r_1 \neq r_1 + 1$r1​=r1​+1, we must use the property $^{n}C_{x} = ^{n}C_{y} \Rightarrow x + y = n$nCx​=nCy​⇒x+y=n:

$$r_1 + (r_1 + 1) = n$$r1​+(r1​+1)=n

$$2r_1 + 1 = n$$2r1​+1=n

Substituting $r_1 = \dfrac{3n + 7}{7}$r1​=73n+7​:

$$2\left(\dfrac{3n + 7}{7}\right) + 1 = n$$2(73n+7​)+1=n

$$\dfrac{6n + 14}{7} + 1 = n$$76n+14​+1=n

$$6n + 14 + 7 = 7n$$6n+14+7=7n

$$n = 21$$n=21

Answer: $21$21