JEE Main 2022 — Binomial Theorem Question with Solution

Given,

$$(5+x{)}^{500}+x(5+x{)}^{499}+x^2(5+x{)}^{498}+$$ ...... $$x^{500}$$

This is a G.P. with first term $$(5+x{)}^{500}$$

Common ratio $$=\frac{x{(5+x)}^{499}}{{(5+x)}^{500}}=\frac{x}{5+x}$$ and 501 terms present.

$$\therefore$$ Sum $$=\frac{{(5+x)}^{500}\left({\left(\frac{x}{5+x}\right)}^{501}-1\right)}{\frac{x}{5+x}-1}$$

$$=\frac{{(5+x)}^{500}\left(\frac{x^{501}-{(5+x)}^{501}}{{(5+x)}^{501}}\right)}{\frac{x-5-x}{5+x}}$$

$$=\frac{\frac{x^{501}-{(5+x)}^{501}}{5+x}}{\frac{-5}{5+x}}$$

$$=\frac{1}{5}\left({\left(5+x\right)}^{501}-x^{501}\right)$$

Coefficient of x¹⁰¹ in $$(5+x{)}^{501}$$ is $$={}^{501}{C}_{101}.5^{400}$$

$$\therefore$$ In $$\frac{1}{5}\left({(5+x)}^{500}-x^{501}\right)$$ coefficient of x¹⁰¹ is $$=\frac{1}{5}.{}^{501}{C}_{101}.5^{400}$$

$$={}^{501}{C}_{101}.5^{399}$$