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Current time:0:00Total duration:6:55

so we're asked to find the sum of the first 50 terms of this series and you might immediately recognize that is a geometric series when we go from one term to the next what are we doing well we're multiplying by 10 11 s to go from 1 to 10 11 s you multiply by 10 over 11 s then you multiply by 10 over 11 so 10 11 10 over 11 again and we keep doing this and we want to find the first 50 terms of it so we can apply the formula we derived for the sum of a finite geometric series and that tells us that the sum of let's say in this case the first 50 terms actually let me do it down here so the sum the sum of the first 50 terms is going to be equal to the first term which is 1 so it's going to be 1 times 1 minus let me do that in a different color 1 times 1 minus the common ratio so the common ratio here is 10 11 10 11 to the 50th power to the power of how many terms we have all of that over 1 minus our common ratio and so I'm not going to solve it completely but we can simplify this a little bit this is going to be 1 minus let me put parentheses here just to make sure we're not just taking the 10 to the 50th power so 1 minus 10 11th to the 50th power over this is 11 11 minus 10 11 s is over 1 over 11 and so this is the same thing as multiplying the numerator by 11 so this is going to be equal to 11 times 1 minus 10 11 s to the 50th power and you could try to simplify this even more but this gets us pretty far at this point it is just arithmetic let's do another one of these this is kind of fun so this is more clearly a geometric series and let's just first think about how many terms we're going to take the some of you might be tempted to say okay I'm going to taking to the 79th power there must be 79 terms here but be very careful because the first term is when we're taking things to the zeroth power we're taking 0.99 to the zeroth power the the second term is where we're taking it to the first power the third term is where we're taking it to the second power the fourth term is where we're taking it to the third power so on and so forth so this right over here is the 80th the 80th term 80th term so we want to find s sub 80 and so this is going to be equal to our first our first term it's going to be 1 times 1 minus our common ratio to the 80th power to the 80th power all over and I'm leaving a blank because we still need to figure out our common ratio all over 1 minus our common ratio so at first you might say well maybe the common ratio here is 0.99 but notice we have a change in sign here and the key thing is to say well to go from one term to the next what are we multiplying by well to go from the first term to the second term we multiply by negative 0.99 and then so we're multiplying by negative 0.99 now to go to the next term we're again multiplying by negative 0.99 so the calm the the common ratio is not positive 0.99 but negative zero negative 0.99 let me write that negative 0.99 and of course that is going to be to the 80th power all over one minus negative 0.99 and so we could simplify this a little bit this is all going to be equal to well that one we don't have to worry too much about that and so this is going to be one minus so negative 0.99 to the 80th power I should put parentheses there to make sure we're we are taking the negative the 0.99 to the 80th power well we're taking it to an even power so it's going to be positive so that's going to be the same thing as 0.99 to the 80th power and all of that over well subtracting a negative that's just going to be adding the positive so all of that over one point nine nine and we could attempt to simplify it more but if we had a calculator we could actually find this this exact value or closed value actually most most calculators don't give you the exact value and you take something to the 80th power but this is this is what that sum is going to be let's do one more of these all right so here we have a series defined recursively and so it's useful to just think about what it would actually look like so the first term is 10 and then the next term so the second term a sub 2 is equal to a sub 1 times 9/10 all right so the next term is going to be the previous term times 9 10 so it's going to be 10 times 9 over 10 and then the next term is going to be that times it's going to be the second term the third term is the second term times 9/10 so 10 times 9 over 10 9 over 10 squared and the way it's written right now we don't I haven't written it as a finite geometric series so let's say we want to take the sum let's say we want the sum of first first I don't know 30 terms sum of first 30 terms so what will this be well we're going to take s sub 1 s sub 30 I know I wrote 10 s sub 30 the sum of the first 30 terms is going to be equal to the first term we've done this before the first term times 1 minus the common ratio 1 minus the common ratio to the 30th power all of that over 1 minus the common ratio and let's see we could 1 minus 9/10 this is 1/10 right over here you divide by 1/10 this the same thing as multiplying by 100 so this is going to be 100 times 1 minus 9/10 to the let me write it this way 9/10 to the 30th power and actually these parentheses you always want to parentheses there just make sure we see they're taking about the 9 and the 10 or the 9/10 the whole thing to the 30th power not just the 9 so there there you go did I - yep there you go we're done