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Q.If (1+x)n=∑r=0n(nr)xr(1 + x)^n = \sum_{r=0}^{n} \binom{n}{r} x^r, then ∑r=0nr2(nr)\sum_{r=0}^{n} r^2 \binom{n}{r} equals

a
n(n−1)2n−2+n2n−1n(n-1)2^{n-2} + n2^{n-1}
b
n(n+1)2n−1n(n+1)2^{n-1}
c
n22n−1n^2 2^{n-1}
d
n(n−1)2n−1n(n-1)2^{n-1}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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