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Q.The sum of first n terms of the series 121+12+221+2+12+22+321+2+3+⋯\frac{1^2}{1} + \frac{1^2+2^2}{1+2} + \frac{1^2+2^2+3^2}{1+2+3} + \cdots is

a
n(n+1)2\frac{n(n+1)}{2}
b
n(n+1)(n+2)6\frac{n(n+1)(n+2)}{6}
c
(n+1)(n+2)2\frac{(n+1)(n+2)}{2}
d
n(n+1)n(n+1)

Correct Answer: Option D

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

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