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

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) We evaluate the triple summation from the inside out:$\sum_{k=1}^j 1 = j$Next,$\sum_{j=1}^i j = \frac{i(i+1)}{2}$Finally,$\sum_{i=1}^n \frac{i(i+1)}{2} = \frac{1}{2} [\sum_{i=1}^n i^2 + \sum_{i=1}^n i]$Using the standard summation formulas $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum i = \frac{n(n+1)}{2}$:$= \frac{1}{2} [\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}]$$= \frac{n(n+1)}{4} [\frac{2n+1}{3} + 1]$$= \frac{n(n+1)}{4} [\frac{2n+4}{3}]$$= \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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