# Confused on sigma notation

1. Jan 8, 2009

### haydn

I understand how to solve sigma notation problems where the index variable is equal to 1, but how would I solve a problem like this??

n
$$\Sigma$$ (i2-3)
i=3

If i was equal to 1 I would be able to solve this, but I'm not sure what to do since it is equal to 3.

Thanks!

2. Jan 9, 2009

### symbolipoint

You want to start the index at i=1 instead of i=3, and you want to know how to change the general term of i2-3.

If you shift the initial index from 3 to 1, then you need to change the general term the same number of index units in the other direction. That is how you compensate for the change to the index variable.

I'm searching for the proper latex formatting in this message tool set but cant' find it.
Summation from 3-2 to n, of (i+(3-1))2-3
You know you wanted to change the starting index value by 2 units to the left, so you change the general term by 2 units to the right.

3. Jan 9, 2009

### symbolipoint

I just need to try this:

$$^{n}_{1}$$$$\sum$$(i+2)2-3

Not exactly the way I hoped it would look, but it is very close; maybe readable by understanding reading members.

Last edited: Jan 9, 2009
4. Jan 9, 2009

### NoMoreExams

Think about it, if we have

$$\sum_{i=1}^{n} i^{2} - 3$$

Now let's write out the first term, we now have

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

Keep writing out the terms until you get to your sum i.e. the one that starts from 3.

Do you see what the answer is?

5. Jan 9, 2009

### HallsofIvy

i starts and 3, it is not "equal to 3". If you were able to do it with i starting at 1, then one way to do this problem is to write out what you would have if i started with 1, then throw away the first two terms!