The sum of the coefficients of ${x}^{499}$ and ${x}^{500}$ in ${\left(1+x\right)}^{1000}+x{\left(1+x\right)}^{999}+{x}^{2}{\left(1+x\right)}^{998}+...+{x}^{1000}$ is: [JEE Main 2026, 28 Jan (Shift 2)]

Question

YESPYQ · Accepted Answer

Correct answer: d) ${}^{1002}{C}_{500}$. Let $S={(1+x)}^{1000}+x{(1+x)}^{999}+{x}^{2}{(1+x)}^{998}+\ldots .+{x}^{1000}$ $={(1+x)}^{1000}\frac{\left(1−{\left(\frac{x}{1+x}\right)}^{1001}\right)}{1−\frac{x}{1+x}}$ $={(1+x)}^{1001}−{x}^{1001}$ The sum of the coefficients of $x^{499}$ and $x^{500}$ ${=}^{1001}{C}_{499}{+}^{1001}{C}_{500}{=}^{1002}{C}_{500}$

The sum of the coefficients of \({x}^{499}\) and \({x}^{500}\) in \({\left(1+x\right)}^{1000}+x{\left(1+x\right)}^{999}+…

Practice more JEE Maths PYQs