###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Geometric Series - Problem 1

Carl Horowitz
###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Finding the sum of a sequence, whenever we're finding a sum of a sequence, the first thing we have to do is identify what kind of sequence it is, what kind of series when you're adding.

So what we need to do is we're looking at this series and we're trying to find s sub 7, the sum of the first terms. So we know we have formulas for arithmetic sums and we also have the formulas for geometric sums, we need to figure out which one we can use. So let's looking at this just try to figure out what kind of series it is.

So to go from, I always jump to terms that I actually find easiest to deal with so what I see here is 1/2 to 3/2 I'm actually adding 1, so I added 1 to get to here. For this to be a arithmetic sequence, I would also have to add 1 to get from 1/6 to 1/2, but I know that's not the case, so I know that this is an arithmetic.

So the next thing we have to consider is geometric, how do I get from 1/2 to 3/2? Multiply by 3. To get from 1/6 to 1/2 is that the same thing do I have to multiply by 3? 1/6 times 3 is three-sixth, simplified is 1/2, so that tells me this is a geometric series and I also know that my rate is 3.

So from that what I can do is go to the equation for a geometric series, s sub n is equal to a1 1 minus r to the n over 1 minus r an you may use a different formula you may switch your rs and your 1s, that's perfectly fine, it's the same exact equation.

So now all we have to do is plug in our information, so we are looking for s sub 7, a1 is our first term, 1/6, 1 minus r is 3 to the n which is 7 over 1 minus r so over 1 minus 3.

In general what you actually have to give your teacher as an answer could vary, from my classes this right here is going to give us a big number, it's going to be ugly, I'm not going to worry about how to deal with that. You could plug in your calculator if you wanted to, in general I don't even ask my students to do that, but what I do want them to do is simplify that piece.

The denominator is pretty easily simplified, it's just subtraction so in this case we just have 1 minus 3 which is -2. So we have 1/6, 1 minus 3 to the seventh over -2 dividing by a term in denominator just multiplying by the reciprocal, so what we actually end up with then is the -2 gets multiplied in here leaving us with 1 over -12, 1 minus 3 to the seventh.

So this will be in simplified form for my class, you may need to plug in your calculator, different teachers look for different things, but basically using our geometric series equations to find the sum use your formula just plug in.