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### Course: Calculus, all content (2017 edition)>Unit 7

Lesson 1: Sequences review

# Worked example: sequence recursive formula

Using the recursive formula of a sequence to find its fifth term. Created by Sal Khan.

## Want to join the conversation?

• Is there any way to find something like a20 without solving a3-a19 first?
• Yes. You need to find a pattern that allows you to state what the term will be based on its index number rather than on any previous term. So, you list out the first few terms and then look for the pattern. Sometimes this is quite easily done, sometimes it is very difficult.
• what are the flat out formulas for the explicit formula and recursive formula for arithmetic and then the same but for geometric?
• Arithmetic: First term: a. Difference: d.
Explicit: nth term=a+(n-1)d.
Recursive: nth term=(n-1) term+d.
Geometric: First term: a. Ratio: r.
Explicit: nth term=a(r)^(n-1).
Recursive: nth term=(n-1) term * r.
• How would you find asub4 if you didn't have asub0?
• In this particular video the sequence is like : a0=2,a1=3,a2=6,a3=18,a4=108 .
But their common difference isn't the same like a1-a0=3-2=1 And a2-a1=6-3=2 they are different so how could it be an arithmetic sequence?
(1 vote)
• You are correct - it is not an arithmetic sequence, but was never claimed as one either, just a sequence that was defined recursively. The video demonstrates how to find the nth term of a recursive sequence.
• My teacher taught it so it was An =An-1 + d.
Does that work with this or is it totally different?
• That works with this. Your teacher's equation is just another example of a recursive sequence. That particular one is the recursive definition of an arithmetic sequence (try it out and see).
• Is there an easier way to know the formula?
(1 vote)
• Shouldn't these videos be more part of the algebra playlist? I remember learning about sequences and series in algebra, but I guess I haven't checked out the rest of this playlist, so I could be wrong.
(1 vote)
• :D Sequences and series are covered in depth in AP Calculus. I took AP Calculus BC and the problem in looks like an AP test question.
• Is there a faster way to find A-sub N? On a test, it takes a lot of time.
(1 vote)
• Unfortunately not, if the series is defined recursively, it's usually faster to evaluate the terms individually than trying to get a explicit formula.
• I understand the example in the video, but the problems in the practice section were really hard. How do I do the one that says A0=3 and An=1/(An-1)+1? It wants us to find the A3.
(1 vote)
• Well, recursively mean we need find the term using the previous term. So to find A₃ you need to know what A₂, A₁, and A₀ are. We are given A₀ = 3 and the formula for A_n = 1/(A_(n-1)) + 1
So to find A₁, use the formula, A₁ = 1/(A₁-₁)) + 1 = 1/A₀+1, then A₂=1/(A₂-₁) + 1 = 1/A₁ +1, and A₃=1/(A₃_₁) +1=1/A₂+1. I think you can take it from here by substitute in what A₂, A₁, and A₀ are.
• Find the next two terms of the sequence
2n+1
• Ok, I'm not exactly sure what you are asking for here. I'm going to try to explain based off what I think you are asking.

So, look at this sequence as:

a sub n = 2n + 1
a sub 1 = 2(1) + 1 = 3
a sub 2 = 2(2) + 1 = 5
a sub 3 = 2(3) + 1 = 7
and so on...

I hope this helps you find the answer you are looking for.