If n is a positive integer, the value of (2n+ | vedclass

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Question

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot

Vedclass · Accepted Answer

$(n+1) 2^n$

Vedclass · Answer

(A) The given expression is $S = \sum_{r=0}^n (2n+1-2r) {^nC_r}$.
Expanding the summation:
$S = (2n+1) \sum_{r=0}^n {^nC_r} - 2 \sum_{r=0}^n r \cdot {^nC_r}$.
We know that $\sum_{r=0}^n {^nC_r} = 2^n$ and $\sum_{r=0}^n r \cdot {^nC_r} = n \cdot 2^{n-1}$.
Substituting these values:
$S = (2n+1) \cdot 2^n - 2 \cdot (n \cdot 2^{n-1}) = (2n+1) \cdot 2^n - n \cdot 2^n = (2n+1-n) \cdot 2^n = (n+1) \cdot 2^n$.
Also, if $f(x) = x^{n+1}$, then $f'(x) = (n+1)x^n$, so $f'(2) = (n+1)2^n$.
Thus, both options $A$ and $C$ are correct.

If $n$ is a positive integer, the value of $(2n+1) ^nC_0 + (2n-1) ^nC_1 + (2n-3) ^nC_2 + \ldots + 1 \cdot ^nC_n$ is

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