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AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 10
Lesson 1: Defining convergent and divergent infinite series- Convergent and divergent sequences
- Worked example: sequence convergence/divergence
- Sequence convergence/divergence
- Partial sums intro
- Partial sums: formula for nth term from partial sum
- Partial sums: term value from partial sum
- Partial sums intro
- Infinite series as limit of partial sums
- Partial sums & series
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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)
The partial sum of a sequence gives us the sum of the first n terms in the sequence. If we know the formula for the partial sums of a sequence, we can find the value of any term in the sequence.
Want to join the conversation?
Why are we substracting S sub 6. Why can´t we just plug in 7 for S sub n?(1 vote)
Unfortunately, S_7 would be the sum of the first 7 terms. Normally the S_n formula for a partial sum means... a sum of those first n terms! If you just want ONE single term though, the partial sum formula isn't going to help directly.
But, if you subtract the terms prior to that term, then you get the term alone! For example:
S_4 = a1 + a2 + a3 + 4
S_3 = a1 + a2 + a3
S_4 - S_3 = a1 - a1 + a2 - a2 + a3 - a3 + a4
OR S_4 - S_3 = 0 + 0 + 0 + a4 = a4!
That's the gist of it. If you want the nth term in a series, then you take the nth partial sum, and subtract the n-1 partial sum from it, this gives you simply the nth TERM (not the partial sum, but a term you would add toward the sum).(15 votes)
Why can't we just come up with a rule for a(sub n) and then input 7 into our formula? I tried doing it that way but I ended up with a completely different answer, can you explain why we can't use this method?(4 votes)
In fact, we can come up with a rule for a_n in this example!
All you have to do is to simplify the following expression which corresponds to S(n) - S(n-1):
(n²+1)/(n+1) - ((n-1)²+1)/((n-1)+1)
If you do the steps correctly, you'll end up with this expression:
a_n = (n²+n-2)/(n²+n).
If you try to evaluate a_7, you will indeed get 27/28 as a result, as expected.
Have a nice day!(1 vote)
Does anyone have a video that explains how to get the rule for a_n with S_n?(1 vote)
So a_n = S_n - S_(n-1)(2 votes)
- why doesnt it work to do S_n - S_(n-1) = a_n (you get (1-n)/(n+1) )and then plug in 7? I did it that way and got a wrong answer, but I'm wondering why it doesnt work(0 votes)
- When I subtracted S_(n-1) from S_n, I got (n+2)(n-1)/(n(n+1)). Plugging in 7 for n yields 54/56, same as the video. There may have been a slight mistake with your algebra.(1 vote)
-2 + i + i^2
----------------- the rule i believe
i (1 + i)(0 votes)
Video transcript
- [Instructor] 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 one over n plus one. 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. All right, 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 one,
plus a sub two, I'll just go all the way, a sub
three, plus a sub four, plus a sub five, plus 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. 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 gonna 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, 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. And if you were 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, wherever we see an
n, 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. And from here we just have to
do a little bit of arithmetic. So, this is going to be
seven squared plus one, this is 49 plus one so
that is 50 over eight. And this is six squared plus one, that's 37, over seven. So, we want to find a common denominator between eight and seven. That would be 56. So this is going to be something over 56, minus something else over 56. Now, to go from eight to
56 I multiply by seven, so I need to multiple the
numerator by seven as well. 50 times seven is 350. And then, this second fraction,
I multiple the denominator by eight to get to 56,
so I have to multiply 37 times eight. Let's see 37 times eight 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 54/56. And if we wanted to
reduce this a little bit, before we rewrite it
maybe in a simpler form, we're rewriting the same value. This would be, let's see, could
we write it as 27 over 28? Let's see, is that about? Yep, that's about as
simplified as we can get. There you go, that's what
a sub seven is, it's 27/28. The difference between the
sum of the first seven terms, and the sum of the first six terms.