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

(B) ${ }^{1002} \mathrm{C}_{500}$

(A) ${ }^{1001} \mathrm{C}_{501}$

(C) ${ }^{1002} \mathrm{C}_{501}$

(D) ${ }^{1000} \mathrm{C}_{501}$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Binomial Theorem

The sum of the coefficients of

x^{499}

and

x^{500}

(1 + x)^{1000} + x (1 + x)^{999} + x^{2} (1 + x)^{998} + \dots\dots + x^{1000}

Answer: (B)

The expression is a geometric series with ratio

\frac{x}{1 + x}

, first term

(1 + x)^{1000}

, and

1001

terms:

S = k = 0 \sum 1000 x^{k} (1 + x)^{1000 - k} = (1 + x)^{1000} \cdot \frac{1 - ( \frac{x}{1 + x} ) ^{1001}}{1 - \frac{x}{1 + x}} = (1 + x)^{1001} - x^{1001} .

The

x^{1001}

term contributes nothing to the coefficients of

x^{499}

x^{500}

, so these come from

(1 + x)^{1001}

and equal

(499 1001)

and

(500 1001)

. By Pascal's rule,

(499 1001) + (500 1001) = (500 1002) .

The required sum of coefficients is

^{1002} C_{500}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option