Question

If the coefficients of $r^{th}$ term and $(r+4)^{th}$ term are equal in the expansion of $(1 + x)^{20}$, then the value of r will be

• A. 8
• B. 11
• C. 9
• D. 10

Answer 1

Answer: (C)

9

Answer 2

The general term in $(1+x)^{20}$ is $T_{k+1} = \binom{20}{k}x^k$. The coefficient of the $r^{th}$ term is $\binom{20}{r-1}$ and the coefficient of the $(r+4)^{th}$ term is $\binom{20}{r+3}$.

Equating these coefficients, $\binom{20}{r-1} = \binom{20}{r+3}$. Using the property $\binom{n}{a} = \binom{n}{b} \implies a=b$ or $a+b=n$, we discard $r-1=r+3$ as it's impossible.

So, $r-1+r+3=20$, which simplifies to $2r+2=20$, leading to $2r=18$, and $r=9$.