Q41 / 75 · JEE Main 2026 (28 January Shift 2) · Mathematics
The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\ldots+x^{1000}$ is :
- Correct answerA${ }^{1002} C_{500}$
- B${ }^{1002} C_{501}$
- C${ }^{1001} C_{501}$
- D${ }^{1000} C_{501}$
Solution
$S = (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \cdots + x^{1000}$
This is a geometric series with first term $(1+x)^{1000}$, common ratio $\frac{x}{1+x}$, and 1001 terms.
$S = (1+x)^{1000} \cdot \dfrac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}}$
$= (1+x)^{1001} - x^{1001}$
Required sum = coefficient of $x^{499}$ + coefficient of $x^{500}$ in $(1+x)^{1001} - x^{1001}$
$= {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500}$
This is a geometric series with first term $(1+x)^{1000}$, common ratio $\frac{x}{1+x}$, and 1001 terms.
$S = (1+x)^{1000} \cdot \dfrac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}}$
$= (1+x)^{1001} - x^{1001}$
Required sum = coefficient of $x^{499}$ + coefficient of $x^{500}$ in $(1+x)^{1001} - x^{1001}$
$= {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500}$
Hence, the correct answer is option: 'A'.