Integrals of sum. FInd upper and lower bounds

Question

Find the upper and lower bound using integrals.

$$\sum_{k=1}^n (k^2 - 3k)$$

Please explain I actually want to understand it

Answer 1

A difficulty with this question is that it asked for "the" upper and lower bounds. There are many upper and lower bounds that could be used.

In general, if $f(x)$ is increasing, $\int_{k-1}^{k} f(x) dx < f(k) <\int_k^{k+1} f(x) dx $, with the inequalities being reversed if $f(x)$ is decreasing.

In your case $f(x) = x^2-3x$ is decreasing at $x=1$ and increasing for $x \ge 2$. This can be handled by separating out the terms for $k=1$ and $2$, writing $\sum_{k=1}^n (k^2 - 3k) = -2+\sum_{k=2}^n (k^2 - 3k) = -2-2+\sum_{k=3}^n (k^2 - 3k) $ and bounding the second or third sum, the terms of which are increasing.