# Thread: really confused with sequences, need some help

1. ## really confused with sequences, need some help

hi all,

we had a "substitute professor" for our discrete class the other day, and to say the least she did a horrible job. I really do not understand how you solve this question at all. Any pointers will be very greatly appreciated. I don't even know where to start.

2. ## Re: really confused with sequences, need some help

You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?

Use the same process for (b) through (d).

3. ## Re: really confused with sequences, need some help

Originally Posted by SlipEternal
You find it one number at a time. For (a), for example, find $a_1$ by plugging in $1$ everywhere you see an $n$. So, $a_1 = 7a_{1-1}-5 = 7a_0-5$. You are given that $a_0 = 1$, so $a_1 = 7(1)-5 = 2$. Then plug in $2$ for $n$: $a_2 = 7a_{2-1} - 5 = 7a_1-5 = 7(2) - 5 = 9$. Can you figure out $a_3, a_4, a_5$ from there?

Use the same process for (b) through (d).

I believe I understand it, so, for a3 = 7a3-1 - 5 = 7a2-5 = 7(9) - 5 = 58

would that be correct?

4. ## Re: really confused with sequences, need some help

Originally Posted by michaelgg13

10. ## Re: really confused with sequences, need some help

Hello, michaelgg13!

Perhaps you are confused by the notation?

$\text{Find }a_5\text{ given: }\:a_n \:=\:7a_{n-2} - 5,\;a_0 = 1$

We have a sequence of numbers: . $a_0,\,a_1,\,a_2,\,a_3,\,\hdots$

It begins with $a_0 = 1$.

The succesive terms are given by: . $a_n \:=\:7a_{n-1}-5$

This means: $\text{the }n^{th}\text{ term is "7 times the previous term, minus 5".}$

$\begin{array}{cccccccc}\text{So that:} & a_0 &=& 1 \\ & a_1 &=& 7(1) - 5 &=& 2 \\ & a_2 &=& 7(2)-5 &=& 9 \\ & a_3 &=& 7(9)-5 &=& 58 \\ & a_4 &=& 7(58)-5 &=& 401 \\ & a_5 &=& 7(401)-5 &=& 2802 \end{array}$

Get the idea?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\text{FYI: }\,\text{The }n^{th}\text{ term is: }\:a_n \;=\;\tfrac{1}{6}(7^n+5)$