If (1+x)^n = sum_{r=0}^n C_r x^r,then the val | vedclass

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(D) The given expression is $S = \sum_{k=0}^n \sum_{r=0}^k C_r$. By changing the order of summation,we have $S = \sum_{r=0}^n \sum_{k=r}^n C_r$. This

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$(n+2) 2^{n-1}$

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(D) The given expression is $S = \sum_{k=0}^n \sum_{r=0}^k C_r$. By changing the order of summation,we have $S = \sum_{r=0}^n \sum_{k=r}^n C_r$. This simplifies to $S = \sum_{r=0}^n C_r (n - r + 1)$. Expanding this,we get $S = (n+1) \sum_{r=0}^n C_r - \sum_{r=0}^n r C_r$. We know that $\sum_{r=0}^n C_r = 2^n$ and $\sum_{r=0}^n r C_r = n 2^{n-1}$. Substituting these values,$S = (n+1) 2^n - n 2^{n-1}$. $S = 2n 2^{n-1} + 2^n - n 2^{n-1} = n 2^{n-1} + 2^n = (n+2) 2^{n-1}$.

If $(1+x)^n = \sum_{r=0}^n C_r x^r$,then the value of $C_0 + (C_0 + C_1) + (C_0 + C_1 + C_2) + \ldots + (C_0 + C_1 + C_2 + \ldots + C_n)$ is

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