If (1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C | vedclass

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(B) Given $(1+x)^n = \sum_{r=0}^n C_r x^r$.We know that $C_r = C_{n-r}$.The expression is $S = C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$.This is t

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$\frac{(2n)!}{(n-2)!(n+2)!}$

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(B) Given $(1+x)^n = \sum_{r=0}^n C_r x^r$.We know that $C_r = C_{n-r}$.The expression is $S = C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$.This is the coefficient of $x^{n-2}$ in the product $(1+x)^n (1+x)^n = (1+x)^{2n}$.Alternatively,it is the coefficient of $x^2$ in $(1+x)^n (1+1/x)^n = \frac{(1+x)^{2n}}{x^n}$.This is the coefficient of $x^{n+2}$ in $(1+x)^{2n}$,which is $^{2n}C_{n+2}$.$^{2n}C_{n+2} = \frac{(2n)!}{(n+2)!(2n-(n+2))!} = \frac{(2n)!}{(n+2)!(n-2)!}$.

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n$,then $C_0C_2 + C_1C_3 + C_2C_4 + .... + C_{n-2}C_n$ equals

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