JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is

A
$^{1001} C_{501}$
B
$^{1002} C_{500}$
C
$^{1002} C_{501}$
D
$^{1000} C_{501}$

Solution

The series $S = (1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots + x^{1000}$ is a geometric progression in the ratio $\frac{x}{1 + x}$ .

Summing the GP: $S = (1 + x)^{1000} \cdot \frac{1 - ( \frac{x}{1 + x} ) ^{1001}}{1 - \frac{x}{1 + x}} = (1 + x)^{1001} - x^{1001}$ .

Coefficient of $x^{499}$ in $S$ is $(499 1001)$ and coefficient of $x^{500}$ is $(500 1001)$ .

Required sum $= (499 1001) + (500 1001) = (500 1002)$ by Pascal's identity.

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