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 = 101,a50,b50 equals:

# ${x}^{n}={a}_{0}+{a}_{1}\left(1+x\right)+{a}_{2}\left(1+x{\right)}^{2}+\dots +{a}_{n}$$\left(1+x{\right)}^{n}$$={b}_{0}+{b}_{1}\left(1-x\right)+{b}_{2}\left(1-x{\right)}^{2}+\dots +{b}_{n}\left(1-x{\right)}^{n}$ then for $\left({a}_{50},{b}_{50}\right)$ equals:

1. A

$\left({-}^{101}{C}_{50}{,}^{101}{C}_{50}\right)$

2. B

3. C

$\left({-}^{101}{C}_{50},{-}^{101}{C}_{50}\right)$

4. D

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### 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 in

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

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

For $n=101,k=51$ we get