# sigma notation

• Nov 3rd 2007, 08:52 AM
Tom123
sigma notation
Hi.

I have this example in my boo

I have a similar question which I started,
But I don't really understand how the sigma notation worked in the first link... therefore I don't know the next line on mine.

Looking for any pointers/help to this next line

Many thanks

tom
• Nov 3rd 2007, 09:01 AM
topsquark
Quote:

Originally Posted by Tom123
Hi.

I have this example in my book http://img229.imageshack.us/img229/686/math1tm3.jpg

I have a similar question which I started, ImageShack - Hosting :: math2um1.jpg

But I don't really understand how the sigma notation worked in the first link... therefore I don't know the next line on mine.

Looking for any pointers/help to this next line

Many thanks

tom

Let's take the first example step by step.
$\sum_{i = 1}^n2^{n - i}(i)2^i$

$= \sum_{i = 1}^n2^{n - i}2^i(i)$

$= \sum_{i = 1}^n2^{n - i + i}(i)$

$= \sum_{i = 1}^n2^n(i)$

$= 2^n \sum_{i = 1}^n(i)$ <-- The $2^n$ does not depend on i so it is constant with respect to the summation

$= 2^n \cdot \frac{n(n + 1)}{2}$

-Dan
• Nov 3rd 2007, 09:12 AM
Tom123
Alright, so the step is simply this?

$4^n$ n(n+1)
........4
• Nov 3rd 2007, 09:33 AM
topsquark
Quote:

Originally Posted by Tom123
Alright, so the step is simply this?

$4^n$ n(n+1)
........4

You mean in your problem? Not quite, because
$\sum_{i = 1}^n 4^{n - i}(i) 4^i = 4^n \sum_{i = 1}^ni = 4^n \cdot \frac{n(n + 1)}{2}$

-Dan