Binomial Theorem
Sum of Coefficients — Geometric Series of Binomials
nta_pyq_2026_jan
Grade 11
Question:
The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\cdots+x^{1000}$ is:
${}^{1002}C_{500}$
${}^{1002}C_{501}$
${}^{1001}C_{501}$
${}^{1000}C_{501}$
Step-by-Step Solution
Key Concept: The sum is a GP with 1001 terms, ratio $\frac{x}{1+x}$. $S=(1+x)^{1000}\cdot\frac{1-(x/(1+x))^{1001}}{1-x/(1+x)}=(1+x)^{1001}-x^{1001}$.
$S=(1+x)^{1001}-x^{1001}$. Required sum $={}^{1002}C_{500}$.
Correct Answer: 1