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I am currently taking a course in Discrete Math. The first part of our lesson this week is regarding sequences. I am stuck on formulas like the ones shown in the images I attached... I was hoping someone might be able to help me learn how to solve them. :)

Ps: What does it mean when $n-1$ is written below the function? Is it the inverse of $a^{n-1}$?

I'm sorry if this is a dumb question haha. I've been studying every day for the past 3 weeks and my brain is officially exhausted. Thank-you so much for your time!

Given $a_n = 3 a_{n-1} + 1$ and $a_0 = 2$, compute $a_2$.

Given the recurrence relation $a_n = -2a_{n-1}$ where $a_0 = 1$, find $a_5$.

Community
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Lindsy Shannon
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2 Answers2

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This is a recursive sequence, which means that every term depends on some of the terms before it (in this case it is precisely the term before it). A well known example of a recursive sequence is the fibonacci sequence.

For your #8, you have to find $a_2$. You are given that $a_0=2$ and that $a_n=3a_{n-1}+1$. This holds for all $n$, which means that:

$a_1=3a_0+1$

$a_2=3a_1+1$

$a_3=3a_2+1$

etc

So $a_1=3a_0+1=3*2+1=7$.

Can you calculate $a_2$?

To clarify your other question, $a_{n-1}$ just means "the term before $a_n$", or "the n-1 term in the sequence".

Jake
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Substitute $n$ by any value that you need.

You know that $$a_0=2$$ then with $n=1$, $$a_1=3a_0+1,$$ and with $n=2$, $$a_2=3a_1+1.$$

The rest is up to you.

Note that

$$a_2=3(3\cdot2+1)+1=3^2\cdot2+3+1.$$

Likewise,

$$a_3=3(3(3\cdot2+1)+1)+1=3^3\cdot2+3^2+3+1$$

and by generalizing

$$a_n=3^n\cdot2+3^{n-1}+3^{n-2}+\cdots3+1.$$

Then let $S:=3^{n-1}+3^{n-2}+\cdots3+1$. You notice that

$$3S=3^n+3^{n-1}+\cdots3^2+3=3^n+S-1,$$ so that

$$S=\frac{3^n-1}2.$$

Finally,

$$a_n=3^n\cdot2+\frac{3^n-1}2.$$