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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 $i$?

S.D.
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Haizum Skallah
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    if you look at the second part of each product, you can see the $1, 2, 3, \ldots, n$ pattern. that's how you pick up the factor of $i$ – citedcorpse Aug 01 '13 at 12:20
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    Take it backwards: If you plug in the value $1$ for $i$ in $\frac bn(\frac{ib}n)^2$ you get $\frac bn(\frac{1\cdot b}n)^2=\frac bn(\frac{b}n)^2$, which is the first summand. If you plug in $i=2$, you get the second summand, ..., if you plug in $i=n$, you get the last summand. – Hagen von Eitzen Aug 01 '13 at 12:23

2 Answers2

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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} $.

Constantine
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I would follow the same path as the comments, especially that of @Hagen von Eitzen.

It's easiest to "expand out" the sum. In general you can write the sum as:

$$\sum_{i=1}^n f(i)$$

where $f(i)$ is a function of $i$. In your example

$$f(i) = \frac{b}{n}\left( \frac{ib}{n} \right)^2 _.$$

You can write $\sum_{i=1}^n f(i)$ as $f(1)$, plus $f(2)$, plus $f(3)$, and so on upto $f(n)$.

$$\sum_{i=1}^n f(i) = f(1) + f(2) + f(3) + \ldots + f(n)$$

In your example

$$\sum_{i=1}^n \frac{b}{n}\left( \frac{ib}{n} \right)^2 = \frac{b}{n}\left( \frac{1b}{n} \right)^2 + \frac{b}{n}\left( \frac{2b}{n} \right)^2 + \frac{b}{n}\left( \frac{3b}{n} \right)^2 + \ldots + \frac{b}{n}\left( \frac{nb}{n} \right)^2$$

It's worth noting that you can take out a common factor from this sum, in just the same way as you can write $ax + ay = a(x+y)$ in elementary algebra. In your case, starting off by repeating the line above, you can take out a common factor of $\frac{b}{n}\left(\frac{b}{n}\right)^2$:

$$\begin{eqnarray*} \sum_{i=1}^n \frac{b}{n}\left( \frac{ib}{n} \right)^2 &=& \frac{b}{n}\left( \frac{1b}{n} \right)^2 + \frac{b}{n}\left( \frac{2b}{n} \right)^2 + \frac{b}{n}\left( \frac{3b}{n} \right)^2 + \ldots + \frac{b}{n}\left( \frac{nb}{n} \right)^2 \\ &=& \frac{b}{n}\left(\frac{b}{n}\right)^2 \left[ 1^2 + 2^2 + 3^2 + \ldots + n^2 \right] \\ &=& \left(\frac{b}{n}\right)^3 \sum_{i=1}^n i^2 \end{eqnarray*}$$

and you end up with getting a much simpler sum. With experience it is possible to pull out common factors without writing out the sum long-hand.

(You can then use a formula to evaluate this simpler sum.)

TooTone
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