We have
$\color{green}{a.}$ $$\left(\frac{b}{n}\right)\left(\frac{b}{n}\right)^2+\left(\frac{b}{n}\right)\left(\frac{2b}{n}\right)^2+\left(\frac{b}{n}\right)\left(\frac{3b}{n}\right)^2+\cdots+\left(\frac{b}{n}\right)\left(\frac{nb}{n}\right)^2 $$
$\color{green}{b.}$ $$=\sum\limits_{i=1}^n\frac{b}{n}\left({\color{red}{{\frac{ib}{n}}}}\right)^2$$
So, in order to make this long and difficult sum more accessible we can put it in "summation notation", by using the sigma symbol.
As we can see in b, what we are trying to say is that b represents the sum of all the summands from the first term ($i=1$) to the nth term ($i=n$) of the common terms. Meaning $b/n$ and something else.
Can some one explain the last part? The origin of term in red $$\left(\frac{ib}{n}\right)^2$$ is not clear for me. How we need to operate the $i$?
Best Answer
You can see that in a second squared term there is another natural (in this case) term $i$, which is increasing by one with each summand. Also, what you have written can be written like this: $(\frac {b}{n})^3\sum_{i=1}^{n}i^2$, because $ b $ and $ n $ are constant (actually, because of the distributive law of multiplication).
EDIT: perhaps this recursive definition for sigma notation will help:
$\sum_{i=a}^{b}f (i) \overset{\underset{\mathrm{def}}{}}{=} \begin{cases} &0 \text{ if } b <a \\ & f (a)+\sum_{i=a+1}^{b}f (i)\text{ if } b\geq a \end{cases}$
Also - $ a, b\in \mathbb {Z} $.