Binomial Theorem Part 3

Concept:The given series is a geometric progression (GP) which simplifies to a single binomial expansion $(1+x)^{1001} - x^{1001}$.The coefficients are found using binomial theorem and Pascal's identity.Explanation:The series is: $S = (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \dots + x^{1000}$.This is a GP with first term $a = (1+x)^{1000}$, common ratio $r = \frac{x}{1+x}$, and $n = 1001$ terms.Using GP sum formula $S = a \frac{1 - r^n}{1 - r}$:$S = (1+x)^{1000} \cdot \frac{1 - \left(\frac{x}{1+x}\right)^{1001}}{1 - \frac{x}{1+x}}$.Simplify denominator: $1 - \frac{x}{1+x} = \frac{1}{1+x}$.Thus $S = (1+x)^{1000} \cdot (1+x) \left[1 - \frac{x^{1001}}{(1+x)^{1001}}\right] = (1+x)^{1001} - x^{1001}$.Now find coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1001}$ (the $-x^{1001}$ term only affects power $1001$).Coefficient of $x^{499} = {}^{1001}C_{499}$.Coefficient of $x^{500} = {}^{1001}C_{500}$.Their sum is ${}^{1001}C_{499} + {}^{1001}C_{500}$.Using Pascal's identity: ${}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1}$, with $n=1001$, $r=499$, we get ${}^{1002}C_{500}$.Answer:${}^{1002}C_{500}$ (Option D).