JEE Main 2026 — 28 January, Evening Shift — Question 2

Let $S = (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \cdots + x^{1000}$. This is a geometric series with first term $a = (1+x)^{1000}$, common ratio $r = \frac{x}{1+x}$, and number of terms $n = 1001$. $S = (1+x)^{1000} \cdot \frac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}} = (1+x)^{1000} \cdot \frac{1 - \frac{x^{1001}}{(1+x)^{1001}}}{\frac{1}{1+x}} = (1+x)^{1001} - x^{1001}$. The coefficient of $x^{499}$ in $S$ is $\binom{1001}{499}$ from $(1+x)^{1001}$ (since $-x^{1001}$ contributes no $x^{499}$). The coefficient of $x^{500}$ in $S$ is $\binom{1001}{500}$ from $(1+x)^{1001}$. Sum of coefficients = $\binom{1001}{499} + \binom{1001}{500} = \binom{1002}{500}$ by Pascal's identity. Thus the correct option is B.