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Geometric Series - Problem 3 4,916 views
Evaluating something in summation notation, so summation notation tells us we are starting at whatever bottom term is going to our top term and then adding whatever we get when we plug those terms in .
So by looking at this I'm trying to evaluate a sum I know that much but I don't know what kind of sum it is. It could be a geometric, it could be an arithmetic, or it could be no sort of set thing at all. So what we have to do is just write out the first couple of terms so we see if we can get a pattern.
So for this particular one what we're doing is we're plugging i starting at 2 into this equation and going up to 7, so let's try that out. So we get 8 times 1/2 squared for our first term, then we add what happens when we go to our next term up, so we go up to 3, 8 times 1/2 to the third and continue up 8 times 1/2 to the fourth so on and so forth.
We could multiply these out, but there's really no reason to because what I'm looking for is to see what kind of pattern I'm dealing with, am I adding something to go from term to term or am I multiplying?
Leaving and multiply that is pretty easy to see that to go from 8 times 1/2 squared to 8 times 1/2 cubed, all I have to do is multiply by 1/2, the same thing from going from 8 times 1/2 times 8 to the 1/2 to the fourth, we just multiply by 1/2 which tells me two things, it tells me we have a geometric series and that we have r is equal to 1/2, so by writing out the first couple of terms, I was able to find some pretty valuable information.
So I know that we have a geometric series, so from that we have a equation to sum these up s sub n is equal to a1, 1 minus r to the n over 1 minus r. You may have a different equation, you may be flipping your 1s and your rs, that's perfectly fine it's just a different way of writing the same equation.
Now all we have to do is plug in our information, so a sub 1 is our first term, 8 times 1/2 squared now I want to simplify this up. So 1/2 squared is one-fourth times 8 is 2, so I know that my first term is actually 2. 1 minus r is 1/2 to the n and we're left with 1 minus 1/2.
We need to plug in for this n now, we can't have any variables in our equation, what we need to do is figure out the number of terms we're dealing with. This is a little bit harder let's go back to our initial statement.
So what happens is we first plug in 2, 3 then 4, 5, 6 and 7, so we actually have 6 terms we're evaluating in this series. This one is really easy to count we're only going from 2 to 7, but what if it was 2 to 147, we don't want to count all those out, so how we figure out how many terms are is always the difference between your top and your bottom plus 1, we said they're 6, the difference here is 5 we need to add that 1 in to count because in subtraction we actually don't include both end points, we just include one so we need to add in that one to make it work. So our n is 6 in this case let's go back and plug in our equation. So this is 6 and then depending on your teacher is will vary how much you have to simplify this up.
In general I wont want my students to simplify the coefficient and the denominator but then leave the term in the parenthesis alone. So if I were to do this, I would get one minus 1/2 is just 1/2, 2 divided by 1/2 is just 4, remember we flip then multiply so what this turns out to be is 4 times 1 minus 1/2 to the sixth.
I would be perfectly happy with this, if you want to plug in your calculator to evaluate it you could as well. So what we did is we went from a summation notation, wrote out a couple of terms to see what kind of patterns we could see, once we did that we were able to plug it in into the formula we know for finding geometric series plug in what we can find for our a sub 1 and our r then simplify it.