Answer
See explanation.
Work Step by Step
First three terms:
$(\mathrm{a})\left(\frac{3}{n}+2\right)^{4} \frac{3}{n}+\left(\frac{6}{n}+2\right)^{4} \frac{3}{n}+\left(\frac{9}{n}+2\right)^{4} \frac{3}{n}$
Last two terms:
\[
\left(\frac{3(-1+n)+2}{n}\right)^{4} \frac{3}{n}+\left(\frac{3 n}{n}+2\right)^{4} \frac{3}{n}
\]
We note that the sum starts at $1=k,$ which means that the interval starts from 2, and for the first letter, we add $ \ underline {3} $ to its value.
Observe the sum
(b) $\sum_{k=0}^{n-1}\left(k \cdot \frac{3}{n}+2\right)^{4} \frac{3}{n}$
If we start the sum at $ k = 0 $ and end at $ n-1$, we'll use the left endpoint approximation.
