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# Partial sums: formula for nth term from partial sum

AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.1 (EK)
,
LIM‑7.A.2 (EK)

## Video transcript

partial sum of the series we're going from 1 to infinity summing it up of a sub n is given by and they tell us the formula for the sum of the first n terms and they say write a rule for the for what the actual nth term is going to be now to help us with this let me just create a little visualization here so if I have a sub 1 plus a sub 2 plus a sub 3 and I keep adding all the way to a sub n minus 1 plus a sub n this whole thing this whole thing that I just wrote out that is S sub n this whole thing is s let me this whole thing is S sub n which is equal to n plus 1 over n plus 10 now if I want to figure out a sub n which is the goal of this exercise well I could subtract out the sum of the first n minus 1 terms so I could subtract out this so that is s that is S sub n minus 1 and what would that be equal to well wherever we see an n we'd replace with an N minus 1 so it'd be n minus 1 plus 1 over n minus 1 plus 10 which is equal to n over n plus 9 since you subtract the red stuff from the blue stuff all you're going to be left with is the thing that we want to solve for you're going to be left with a sub n so we could write down a sub n is equal to s sub n is equal to n minus s sub n minus 1 s sub n minus 1 or we could write that is equal to this stuff so this is the n plus 1 over n plus 10 minus minus n over n plus 9 and this by itself this is a rule for a sub n but we could combine these terms add these two fractions together now this is actually going to be the case for N greater than 1 for N equals 1 s sub 1 is going to be well you can just the a sub one is going to be equal to s sub one but then for any other n we could use this right over here and if we want to simplify this well we can add these two fractions we can add these two fractions by having a common denominator so let's see if we multiply the numerator and denominator here times n plus 9 we are going to get so this is equal to n plus 1 times n plus 9 over n plus 10 times n plus 9 and from that we are going to subtract let's multiply the numerator the denominator here by n plus 10 so we have n times n plus 10 over n plus 9 times n plus 10 n plus 9 times n plus 10 and what does that give us so let's see if we simplify up here we're going to have this is N squared plus 10 n plus 9 that's that and then this right over here is N squared plus this is N squared plus 10 n we doing that red color so this is N squared plus 10 n and remember going to subtract this and so we are close to deserving a drumroll a sub n is going to be equal to our denominator right over here is n plus 9 times n plus 10 and we're going to subtract the red stuff from the blue stuff so you subtract an N squared from an N squared those cancel out subtract a 10 n from a 10 n those cancel out and you're just left with that blue 9 so there you have it we have expressed we've written a rule for a sub n for n greater than 1