1+ (1+2) + (1 +2+3) + ..... + (1 +2+3+... +n) =?

A \( \frac{n (n+1)}{2} \)

B \( \frac{n (n+1) (n+2)}{6} \)

C \( \frac{n (2n+1)}{6} \)

D \( \frac{n (4n+1)}{6} \)

Solution

Correct Answer: Option A

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