Question

Prove that ${C_1}^2-2.{C_2}^2+3.{C_3}^2-.....-2n{C_{2n}}^2=(-1{)}^{n-1}n{C}_n.$

Answer 1

$(1+x{)}^{2n}=C_0+C_{1}x+C_2{x}^2+....+C_{2n}{X}^{2n}$....(1)
Differentiating the expansion of $(1+x{)}^{2n}$, we get
$2n(1+x{)}^{2n-1}=C_1+2C_{2}x+3C_3{x}^2+....2n{C}_{2n}{x}^{2n-1}$....(2)
${(1-\frac{1}{x})}^{2n}=C_0-C_1\cdot \frac{1}{x}+C_2\cdot \frac{1}{x^2}-....+C_{2n}\cdot \frac{1}{x^{2n}}$....(3)
where $C_r={}^{2}n{C}_r$
Multiplying (2) and (3), we get
$2n(1+x{)}^{2n-1}{(1-\frac{1}{x})}^{2n}$
$=(C_1+2C_2+3C_3{x}^2+.....+2n{C}_{2n}{x}^{2n-1})\times (C_0-C_1\cdot \frac{1}{x^2}-...+C_{2n}\frac{1}{x^{2n}})$
The coefficient of 1/x in R.H.S.is
$-(C_1^2+2C_2^2+3C_3^2-.....-2n{C}_{2n}^2)$ .....(4)
Also the coefficient of $\frac{1}{x}$ in
$2n(1+x{)}^{2n-1}{(1-\frac{1}{x})}^{2n}$
= coefficients of $\frac{1}{x}$ in $2n\frac{(1+x{)}^{2n-1}(x-1{)}^{2n}}{x^{2n}}$
= coeff. of $\frac{1}{x}$ in $\frac{2n}{x^{2n}}(1-x^2{)}^{2n-1}(1+x)$
= coeff. of $x^{2n-1}in2n(1-x^2{)}^{2n-1}(1-x)$
= coeff. of $x^{2n-2}in2n(1-x^2{)}^{2n-1}\times$ coeff.of x in $(1-x)$
= coeff. of $\left(x^2{)}^{n-1}in2n\right(1-x^2{)}^{2n-1}\times$ coeff.of x in $(1-x)$
$=2n(-1{)}^{n-1}{}^{2n-1}{C}_{n-1}(-1)$
$=(-1{)}^{n}2n\cdot \frac{(2n-1)!}{(n-1)!n!}=(-1{)}^n\frac{\left(2n\right)!}{n\cdot (n-1)!n!}\cdot n$
$=(-1{)}^n\frac{\left(2n\right)!}{n!n!}\cdot n=-(-1{)}^{n-1}n{C}_n;[C_n={}^{2n}{C}_n]$ ..(5)
$\therefore$ from (4) and (5) , we get
$C_1^2-2C_2^2+3C_3^2-.....-2n{C}_{2n}^2=(-1{)}^{n-1}n{C}_n$