Evaluate

after splitting the square term,we have to find

1) S = 2nx\sumnCr r x^r(1-x)^n-r

which may be done by considering the expression P=(1-x + y )ⁿ

if we put x= y in the expression T= ydPdy ,we get S2nx

2)S₁ = \sumnCr r² x^r (1-x)^n-r

S₁ is given by putting x= y in the expression ydTdy

3)S₃ = n²x²\sumnCr x^r (1-x)^n-r

we can have S₃ by putting x=y in P, so S₃ is simply 1

the required term is given by S₁ - S + S₃

\bf\sum_{r=0}^{n}^nC_{r}(r-nx)^2.x^r.(1-x)^{n-r}=nx(1-x)$

\bf(1+y)^n=\sum_{r=0}^{n}^nC_{r}.y^r$\\\\ Now $\bf n.(1+y)^{n-1}=\sum_{r=0}^{n}r.^nC_{r}.y^{r-1}$\\\\ Again $\bf n.(n-1)(1+y)^{n-2}=\sum_{r=0}^{n}r.(r-1)^nC_{r}.y^{r-2}$\\\\ Now $\bf \sum_{r=0}^{n}(r-nx)^2.^nC_{r}.x^r.(1-x)^{n-r}$\\\\ $\bf= \sum_{r=0}^{n}(\underbrace{\bf r^2-r}+\underbrace{\bf r-2r.n.x}+n^2x^2).^nC_{r}.x^r.(1-x)^{n-r}$\\\\ $\bf=(1-x)^n\left\{\sum_{r=0}^{n}^nC_{r}.r(r-1).y^r+(1-2nx).\sum_{r=0}^{n}^nC_{r}.r.y^r+n^2.x^2.\sum_{r=0}^{n}^nC_{r}.y^r \right\}$\\\\ Where $\bf y=\frac{x}{1-x}$\\\\ $\bf=(1-x)^n.\left\{n.(n-1).(1+y)^{n-2}.y^2+(1-2nx).n.(1+y)^{n-1}.y+n^2.x^2.(1+y)^n \right\}$\\\\ After Simplifying, We get.....\\\\ $\bf =nx.(1-x)$\\\\

Hmm. Thannks!