JEE Main 2022 · 25 Jun · Shift 2 · Mathematics Q7

MathematicsSingle correct+4 / -1Question 7 of 90

The coefficient of $x^{101}$ in the expression $(5+x)^{500}+x(5+x)^{499}+x^2(5+x)^{498}+\dots+x^{500}$ , $x>0$ , is

{}^{501}C_{101}(5)^{399}

{}^{501}C_{101}(5)^{400}

{}^{501}C_{100}(5)^{400}

{}^{500}C_{101}(5)^{399}

Answer and worked solution

Official answer: A

Geometric sum. $\displaystyle\sum_{k=0}^{500}x^k(5+x)^{500-k}=\frac{(5+x)^{501}-x^{501}}{(5+x)-x}=\frac{(5+x)^{501}-x^{501}}5$ .

Coefficient of $x^{101}$ . $\dfrac{{}^{501}C_{101}5^{400}}5={}^{501}C_{101}(5)^{399}$ .

(A)

Topics:General and middle terms Geometric progression

JEE Main 2022 · 25 Jun · Shift 2, Mathematics question 7