Question

Prove that [1 + (n + 1) x] $(1+x{)}^{n-1}$
$=C_0+2C_{1}x+3C_2{x}^2+...+(n+1)C_n{x}^n$
and $C_0^2+2C_1^2+3C_2^2+...(n+1)C_n^2$
$=\frac{(n+2)(2n-1)!}{n!(n-1)!}$

Answer 1

(1 + x)${}^n$ = $C_0+C_{1}x+C_2{x}^2+...$
x(1 + x)${}^n$ = $C_{0}x+C_1{x}^2+C_2{x}^3+...$
Diff.both sides
1(1 + x)${}^n$ + n .x(1 + x)${}^{n-1}$ = $C_0+2C_{1}x+3C_2{x}^2+...$
or [1 + (n + 1) x] (1 + x)${}^n$ = $C_0+2C_{1}x+3C_2{x}^2+...$ ...(1)
2nd part : We have to evaluate
$C_0^2+2C_1^2+3C_2^2+.....+(n+1)C_n^2$ ...(2)
consider ${(1-\frac{1}{x})}^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}...$ ...(3)
If we multiply (1) and (3) then (2) occurs as the term independent of x in this product
Hence we have to search the term which is independent of x in
[1 + (n + 1) x](1+ x)${}^{n-1}{(1-\frac{1}{x})}^n$
or$\frac{[1+(n+1\left)x\right](1+x{)}^{2n-1}}{x^n}$
or the coefficients of $x^3$ in the numerator
[1 + (n + 1) x](1+ x)${}^{2n-1}$
or coeff. of $x^n$ in (1 + x)${}^{2n-1}$ + (n + 1)
coeff. of x${}^{n-1}$ in (1 + x )${}^{2n-1}$
$={}^{2n-1}{C}_n+(n+1){\cdot }^{2n-1}{C}_{n-1}$
$\frac{(2n-1)!}{n!(n-1)!}+(n+1)\frac{(2n-1)!}{(n-1)!n!}$
$=\frac{(2n-1)!}{n!(n-1)!}+[1+(n+1\left)\right]=\frac{(n+2)(2n-1)!}{n!(n-1)!}$