JEE Main 2026 Apr 5 Shift 1, Mathematics Q23: Introduction to Binomial Expansion

JEE Main2026Apr 5, Shift 1Mathematics

If the sum of the coefficients of $x^{7}$ and $x^{14}$ in the expansion of $(\frac{1}{x ^{3}} - x^{4})^{n}$ , $x \neq = 0$ , is zero, then the value of $n$ is _______.

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Solution

The general term is $T_{r + 1} = (r n) (x^{- 3})^{n - r} (- x^{4})^{r} = (- 1)^{r} (r n) x^{7 r - 3 n}$ .

For $x^{7}$ : $7 r_{1} - 3 n = 7 \Rightarrow r_{1} = \frac{7 + 3 n}{7}$ . For $x^{14}$ : $7 r_{2} - 3 n = 14 \Rightarrow r_{2} = r_{1} + 1$ .

Sum of coefficients zero: $(- 1)^{r_{1}} (r _{1} n) + (- 1)^{r_{1} + 1} (r _{1} + 1 n) = 0 \Rightarrow (r _{1} n) = (r _{1} + 1 n)$ .

This requires $r_{1} + (r_{1} + 1) = n$ , i.e. $n = 2 r_{1} + 1$ . Substituting $r_{1} = \frac{7 + 3 n}{7}$ :

$n = \frac{14 + 6 n}{7} + 1 \Rightarrow 7 n = 14 + 6 n + 7 \Rightarrow n = 21$ .

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JEE Main 2026 Apr 5 Shift 1, Mathematics Q23: Introduction to Binomial Expansion

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