If C_j = {}^{n}C_j,then C_0 C_r + C_1 C_{r+1} | vedclass

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(C) We know that $(1+x)^{2n} = (1+x)^n (1+x)^n$. Expanding both sides: $(1+x)^{2n} = (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) (C_0 x^n + C_1 x^{n-1}

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$^{2n}C_{n+r}$

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(C) We know that $(1+x)^{2n} = (1+x)^n (1+x)^n$. Expanding both sides: $(1+x)^{2n} = (C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n) (C_0 x^n + C_1 x^{n-1} + C_2 x^{n-2} + \ldots + C_n)$. The coefficient of $x^{n-r}$ in the expansion of $(1+x)^{2n}$ is $^{2n}C_{n-r}$,which is equal to $^{2n}C_{n+r}$. By multiplying the series,the coefficient of $x^{n-r}$ is $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n$. Thus,$C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = {}^{2n}C_{n+r}$.

If $C_j = {}^{n}C_j$,then $C_0 C_r + C_1 C_{r+1} + C_2 C_{r+2} + \ldots + C_{n-r} C_n = $

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