How do I write each sum using sigma notation?

Question

I have a few problems of this for homework, the teacher didn't really explain this very well because this is for extra credit. I only need the steps to do one of these and I would use it as a guide.. please show me the step by step process

a. 625 + 608.25 + 591.5 + 574.75 + .... + 5.25

b. 243 - 81 + 27 - 9 + ...

c. 2.5 + 7.5 + 22.5 +67. 5 + .... + 35.872.267.5

Answer 1

I'll help you with b.

You generally always start out a sigma notation using the Sigma sign, and placing n = 1, or n = 0 below it, and the infinity sign above it.  Like this:

∞

∑

n=1

This statement says that you will be adding all n terms starting at term n and going to infinity.

Because you are given the terms in b, you have a first term (243), so you can use n = 1 in this scenario.

Next you are going to want to figure out what all of the terms in your series have in common.  

You will come to see that all of these numbers are divisible by 3, so you can use that as a base.  Your equation now looks like so:

∞

∑ 3

n=1

*keep in mind that the n = 1 should be completely under the ∑ so that it does not appear as if the 3 is being divided by it.  Sorry, Yahoo! Answers won't let me type it that way.

Now, what do you have to multiply 3 by to get 243?  The answer is a very large number, and that presents a problem because what ever you do to one term in the series you must do to every term.  So, try to think of a number or term that you could raise 3 to, to get 243.

3^5 = 243.  But remember, each term must be multiplied or raised to the same thing.  If that were to happen, all of the terms would be 243, which they are not.  In fact the terms seem to be 3^5, 3^4 = (81), 3^3 = (27), and 3^2 = (9).  This is where your (n) can come in handy.

If you say the terms are 3^(6-n), and you are starting at n=1, that means your first term (n=1) will be 3^(6-1) = 243, your second term (n=2) will be 3^(6-2) = 81, and so on.  Using the variable allows your terms to change, which is what you need.

Now the equation looks like so:

∞

∑ 3^(6-n)

n=1

There is only one thing left to do.  Because your terms are alternating between positive and negative, you need to multiply your sum by a certain number or equation so that your terms can alternate.

These types of series are called "alternating series" and they generally always have this in the beginning:

(-1)^(n+1)

Your equation would then look like this:

∞

∑ (-1)^(n+1) (3)^(6-n)

n=1  

Because you are starting on the first term and it is positive, the (-1)^(n+1) = (-1)^(1+1) = (-1)^2 = 1; your second term (n=2) would then be (-1)^(2+1) = (-1)^(3) = -1, and so on, allowing your terms to alternate.

So,

∞

∑ (-1)^(n+1) (3)^(6-n)

n=1

is your answer.