JEE Main 2014 · 11 Apr · Shift 1 · Mathematics Q67

← JEE Main 2014 · 11 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 67 of 90

The coefficient of $x^{50}$ in the binomial expansion of
$(1+x)^{1000}+x(1+x)^{999}+x^2(1+x)^{998}+\dots+x^{1000}$
is :

\dfrac{(1000)!}{(50)!\,(950)!}

\dfrac{(1000)!}{(49)!\,(951)!}

\dfrac{(1001)!}{(51)!\,(950)!}

\dfrac{(1001)!}{(50)!\,(951)!}

Answer and worked solution

Official answer: D

This is a geometric sum with ratio $\frac x{1+x}$ :
$\sum_{k=0}^{1000}x^k(1+x)^{1000-k}=\frac{(1+x)^{1001}-x^{1001}}{(1+x)-x}=(1+x)^{1001}-x^{1001}.$
The coefficient of $x^{50}$ is $\binom{1001}{50}=\dfrac{(1001)!}{(50)!\,(951)!}$ .

Topics:Properties of binomial coefficients Geometric progression

← Previous question Next question →

JEE Main 2014 · 11 Apr · Shift 1, Mathematics question 67