sumlimits_{i=1}^n sumlimits_{j=1}^i sumlimits | vedclass

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(D) We evaluate the triple summation step by step:$\sum_{k=1}^j 1 = j$Next,$\sum_{j=1}^i j = \frac{i(i+1)}{2}$Finally,$\sum_{i=1}^n \frac{i(i+1)}{2} =

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We evaluate the triple summation step by step:$\sum_{k=1}^j 1 = j$Next,$\sum_{j=1}^i j = \frac{i(i+1)}{2}$Finally,$\sum_{i=1}^n \frac{i(i+1)}{2} = \frac{1}{2} \left[ \sum_{i=1}^n i^2 + \sum_{i=1}^n i \right]$$= \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$$= \frac{n(n+1)}{4} \left[ \frac{2n+1}{3} + 1 \right]$$= \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right]$$= \frac{n(n+1) \cdot 2(n+2)}{12} = \frac{n(n+1)(n+2)}{6}$

$\sum\limits_{i=1}^n \sum\limits_{j=1}^i \sum\limits_{k=1}^j 1 = \dots$

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