sumlimits_{r = 1}^n {sumlimits_{m = 1}^r {m} | vedclass

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(B) We know that the sum of the first $r$ natural numbers is $\sum_{m=1}^r m = \frac{r(r+1)}{2}$.Therefore,the given expression is $\sum_{r=1}^n \frac

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

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(B) We know that the sum of the first $r$ natural numbers is $\sum_{m=1}^r m = \frac{r(r+1)}{2}$.Therefore,the given expression is $\sum_{r=1}^n \frac{r(r+1)}{2} = \frac{1}{2} \sum_{r=1}^n (r^2 + r)$.Using the standard summation formulas $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r = \frac{n(n+1)}{2}$,we get:$= \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}$.

$\sum\limits_{r = 1}^n {\sum\limits_{m = 1}^r {m} } = \dots$

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