# Doubt from Binomial

What a coincidence. I am solving the same paper. Though i have not solved it myself but i have solution

This is out of syllabus. From theory of finite differences numerator =0

Ok sir

Okay Thanks Sir

As Bhatt Sir mention..We can solved it using Finite Diff. Method.

Using \displaystyle f(x)=(1-x)^{n}=\sum^{n}_{k=0}(-1)^{k}\binom{n}{k}\cdot x^{k}.

Diff. Both side w. r. to x

\displaystyle f'(x)=-n(1-x)^{n-1}=\sum^{n}_{k=0}(-1)^{k}\binom{n}{k}\cdot k\cdot x^{k-1}.

Here we have Not that f'(1)=0

Multiply both side by x and Then Diff. Both side w.r.to x

\displaystyle f''(x)=n(1-x)^{n-1}+n(n-1)(1-x)^{n-2}=\sum^{n}_{k=0}(-1)^k\binom{n}{k}\cdot k^2\cdot x^{k-1}.

Here note that f''(1)=0

Again multiply by x and Diff. Both side w. r. to x

We can comtinue this process for higher values of m.

Note that L.H.S is always =0 when x=1 Since we are just taking the derivative each time.

So we have \displaystyle \sum^{n}_{k=0}(-1)^{k}\binom{n}{k}\cdot k^m=0 for m<n.