Binomial Theorem – Sum of Coefficients – JEE Main 28 Jan 2026 Shift 2

Solution:

Let the given series be $S$.
$$S = (1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+……+x^{1000}$$

This is a Geometric Progression (G.P.) with:

First term $a = (1+x)^{1000}$

Common ratio $r = \frac{x}{1+x}$

Number of terms $n = 1001$

Using the sum formula $S_n = a \frac{1-r^n}{1-r}$:
$$S = (1+x)^{1000} \cdot \frac{1 – (\frac{x}{1+x})^{1001}}{1 – \frac{x}{1+x}}$$

Simplifying the denominator:
$$1 – \frac{x}{1+x} = \frac{1+x-x}{1+x} = \frac{1}{1+x}$$

Substituting back:
$$S = (1+x)^{1000} \cdot (1+x) \cdot \left[ 1 – \frac{x^{1001}}{(1+x)^{1001}} \right]$$
$$S = (1+x)^{1001} \left[ 1 – \frac{x^{1001}}{(1+x)^{1001}} \right]$$
$$S = (1+x)^{1001} – x^{1001}$$

We need the sum of the coefficients of $x^{499}$ and $x^{500}$.

Coefficient of $x^{499}$ in $(1+x)^{1001}$ is ${}^{1001}C_{499}$.

Coefficient of $x^{500}$ in $(1+x)^{1001}$ is ${}^{1001}C_{500}$.

Note: The term $-x^{1001}$ does not affect these coefficients.

Required Sum = ${}^{1001}C_{499} + {}^{1001}C_{500}$

Using the property ${}^{n}C_{r-1} + {}^{n}C_{r} = {}^{n+1}C_{r}$:
$${}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500}$$

Ans. (2)