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Q.The sum ∑k=1n(k2+2k+2)\sum_{k=1}^{n} (k^2 + 2k + 2) equals

a
n(n+1)(n+2)3+2n\dfrac{n(n+1)(n+2)}{3} + 2n
b
n(n+1)(2n+1)6+n(n+1)+2n\dfrac{n(n+1)(2n+1)}{6} + n(n+1) + 2n
c
n(n+1)(n+2)3+n(n+1)\dfrac{n(n+1)(n+2)}{3} + n(n+1)
d
n(n+1)(n+2)(n+3)12\dfrac{n(n+1)(n+2)(n+3)}{12}

Correct Answer: Option B

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

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