Question

Prove that: ${\sum }_{r=0}^{n}r(n-r)C_r^2=n^2{(}^{2n-2}{C}_n)$

Answer 1

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+......+C_r{x}^r+.....+C_n{x}^n$ .....(1)
Now the term r(n - r) $C_r^2$ suggests the factor
r($C_r$) (n - r) $C_r$
By complementary rule $C_r=C_{n-r}$
Against r$C_r$ suggest differentiation. The second factor suggest that (1) should be written as
$(x+1{)}^n=C_0{x}^n+C_1{x}^{n-1}+.....+C_r{x}^{n-r}+....+C_n$ .....(2)
and the second factor (n - r)$C_r$ suggest again differentiation of (2).
Differentiate both (1) and (2)
$n(1+x{)}^{n-1}=C_1+2C_{2}x+......+r{C}_r{x}^{r-1}+.....$ ....(3)
$n(1+x{)}^{n-1}=n{C}_0{x}^{n-1}+(n-1)C_1{x}^{n-2}+......+(n-r)C_r{x}^{n-r-1}+......$ ...(4)
Multiplying (3) and (4), we get
$n^2(1+x{)}^{2n-2}=\sum (r{C}_r\left)\right(n-r)C_r{x}^{r-1}{x}^{n-r-1}$
$=\sum r(n-r)C_r^2{x}^{n-2}$
Hence we should search the coefficient of $x^{n-2}$ in L.H.S. and it will be $n^2\cdot {}^{2n-2}{C}_{n-2}=n^2{(}^{2n-2}{C}_n)$ by complementary rule