Partial sums: term value from partial sum | Series | AP Calculus BC | Khan Academy

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. 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 2, I'll just go all the way: a sub 3 plus a sub 4 plus a sub 5 plus a sub 6 plus a sub 7.

So, if I were to sum all of these together, that this entire sum 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 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. 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 this going to be? Well, S sub seven, the sum of the first seven terms, we just—whatever we see in, 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 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. 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 multiplying 37 times 8, and see, 37 times 8 is going to be 240 plus 56, so that is 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 over 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 a new value; it's, we're rewriting the same value.

This would be, we read it as 27 over 28. And let's see, is that about—yep, that's about as simplified as we can get. But there you go, that's what a sub 7 is. It's 27 over 28, the difference between the sum of the first 7 terms and the sum of the first 6 terms.