JEE Main 2026 — 5 April, Morning Shift — Question 46

$T_{r+1}={ }^{n} C_{r}\left(\frac{1}{x^{3}}\right)^{n-r}\left(-x^{4}\right)^{r}$ $\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{7 \mathrm{r}-3 \mathrm{n}}(-1)^{\mathrm{r}}$ for coeff. of $x^{7} \Rightarrow 7 r_{1}-3 n=7$ $\Rightarrow \mathrm{r}_{1}=\frac{7+3 \mathrm{n}}{7}$ coeff. of $x^{7} \Rightarrow{ }^{n} C_{r_{1}}(-1)^{r_{1}}$ for coeff. of $x^{14} \Rightarrow 7 r_{2}-3 n=14$

$$\Rightarrow \mathrm{r}_{2}=\frac{14+3 \mathrm{n}}{7}=\mathrm{r}_{1}+1$$

coeff. of $x^{14} \Rightarrow{ }^{n} C_{r_{1}+1}(-1)^{r_{1}+1}$ $\Rightarrow{ }^{n} C_{r_{1}}(-1)^{r_{1}}+{ }^{n} C_{r_{1}+1}(-1)^{r_{1}+1}=0$ $\Rightarrow{ }^{n} \mathrm{C}_{\mathrm{r}_{1}}={ }^{n} \mathrm{C}_{\mathrm{r}_{1}+1} \Rightarrow \mathrm{r}_{1}+\mathrm{r}_{1}+1=\mathrm{n}$ $\Rightarrow 2 \mathrm{r}_{1}+1=\mathrm{n} \Rightarrow 2\left(\frac{7+3 \mathrm{n}}{7}\right)+1=\mathrm{n}$

Ans. $\Rightarrow \mathrm{n}=21$