If xn=a0+a1(1+x)+a2(1+x)2+………..+an(1+x)n=b0+b1(1−x)+b2(1−x)2+…+bn(1−x)n   then for n=201,a101,b101 is equal to:

# If ${x}^{n}={a}_{0}+{a}_{1}\left(1+x\right)+{a}_{2}\left(1+x{\right)}^{2}+\dots$……..$+{a}_{n}\left(1+x{\right)}^{n}$then for $n=201,\left({a}_{101},{b}_{101}\right)$ is equal to:

1. A

$\left({-}^{201}{C}_{101},{-}^{201}{C}_{101}\right)$

2. B

3. C

$\left({-}^{201}{C}_{101}{,}^{201}{C}_{101}\right)$

4. D

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\begin{array}{l}{x}^{n}=\left[\left(1+x\right)-1{\right]}^{n}=\left[1-\left(1-x\right){\right]}^{n}\\ =\sum _{k=0}^{n}{ }^{n}{C}_{k}\left(1+x{\right)}^{n-k}\left(-1{\right)}^{k}\\ =\sum _{k=0}^{n}{ }^{n}{C}_{k}\left(-1{\right)}^{k}\left(1-x{\right)}^{k}\end{array}$

$\therefore$ ${a}_{k}=$ coefficient of $\left(1+x{\right)}^{k}$ in

$\sum _{k=0}^{n}{ }^{n}{C}_{k}\left(1+x{\right)}^{n-k}\left(-1{\right)}^{k}=\left(-1{\right)}^{n-{k}^{n}}{C}_{n-k}=\left(-1{\right)}^{n-{k}^{n}}{C}_{k}$

and  ${b}_{k}{=}^{n}{C}_{k}\left(-1{\right)}^{k}$

For

we get