If n is a positive integer,then sum_{r=1}^n r | vedclass

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(C) We know that $(1+x)^n = \sum_{r=0}^n C_r x^r = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$. Differentiating both sides with respect to $x$: $n(1+x)^

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$n(n+1)$

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(C) We know that $(1+x)^n = \sum_{r=0}^n C_r x^r = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n$. Differentiating both sides with respect to $x$: $n(1+x)^{n-1} = C_1 + 2C_2 x + 3C_3 x^2 + \ldots + nC_n x^{n-1}$. Multiplying both sides by $x$: $nx(1+x)^{n-1} = C_1 x + 2C_2 x^2 + 3C_3 x^3 + \ldots + nC_n x^n$. Differentiating again with respect to $x$: $n[(1+x)^{n-1} + x(n-1)(1+x)^{n-2}] = C_1 + 2^2 C_2 x + 3^2 C_3 x^2 + \ldots + n^2 C_n x^{n-1}$. Putting $x=1$: $\sum_{r=1}^n r^2 C_r = n[2^{n-1} + (n-1)2^{n-2}] = n[2 \cdot 2^{n-2} + (n-1)2^{n-2}] = n(n+1)2^{n-2}$. Thus,the missing term is $n(n+1)$.

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

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