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Partial sums: term value from partial sum

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

Video transcript

we're told that the nth partial sum of the series from N equals one to infinity of a sub n is given by and so the sum of the first n terms is N squared plus 1 over n plus 1 and they want us to figure out what is the actual seventh term and like always pause this video and see if you can figure it out on your own before we work through it together alright so one way to think about it is a sub seven let's think about how that relates to different sums so if we have a sub 1 plus a sub two I'll just go all the way a sub a sub 3 plus a sub four plus a sub five + a sub six plus a sub seven so if I were to sum all of these things together that this entire sum that would be S sub seven and if I wanted to figure out a sub seven well I could subtract from that I could subtract out the sum of the first six terms so I could sum I could subtract out S sub I could subtract out S sub six so once again what am i doing here what is my strategy I know the formula for the sum of the first n terms I can use that to say okay I can figure out the sum of the first seven terms that's going to be the sum of all of these and then I can use that same formula to figure out the sum of the first six terms and the difference between the two well that's going to be our a sub seven so another way of saying what I just said is that a sub seven is going to be the sum of the first seven terms minus the sum of the first six terms some of the sixth for the sum of the first six terms and if you are doing this problem on your own you wouldn't have to write it out this way I just wrote it out this way hopefully making this statement a little bit more intuitive well what is what are what is this going to be well S sub seven the sum of the first seven terms we just whatever we CNN we replace it with a seven so it's going to be seven squared plus one over seven plus one and from that we are going to subtract S sub six the sum of the first six terms well that's going to be six squared plus one or six plus one over six plus one and from here we just have to do a little bit of arithmetic so this is going to be let's see seven squared plus one this is 49 plus 1 so that is 50 over 8 and this is 6 squared plus one that is 36 plus one that's 37 over seven so let's see we want to find a common denominator between 8 & 7 that would be 56 so this is going to be something over 56 something over 56 minus something else over 56 minus something else over 56 now to go from 8 to 56 I multiply by 7 so I need to multiply the numerator by 7 as well 50 times 7 is 350 and then this second fraction I multiply the denominator by 8 to get 256 so after multiply 37 times 8 and see 37 times 8 is going to be 240 plus 56 so that is 296 296 and so this is going to be equal to so I have a denominator of 56 350 minus 296 is 54 so it's 54 56 and if we wanted to reduce this a little bit before we rewrite it maybe in a simpler form we're not really making us where it's we're rewriting the same value this would be seek we read it as 27 over 28 27 28 and let's see is that about yep that's about as as simplified as we can get but there you go that's what a sub 7 is it's 27 28 the difference between the sum of the first 7 terms and the sum of the first six terms