Simple Sums

Expand the series for each value of $n$ : 3+4+5+6+7+8

Add 3, 4, 5, 6, 7 and 8 to get 33

Since $\displaystyle \sum_{k=3}^8 n$ is a sum of an arithmetic progression, its sum is given by $S = \dfrac {n(a+l)}2$ , where $n = 8 - 3+1 = 6$ is the number of terms, $a = 3$ is the first term, and $l = 8$ is the last term.

Therefore, $\displaystyle \sum_{k=3}^8 n = \frac {6(3+8)}2 = \boxed{33}$ .

Expand it to obtain $3+4+5+6+7+8 = \color{#69047E}{\boxed{33}}$ .