1. ## Sequences

Hi

Could somebody explain how I would go about doing the attached problem. I don't really get it at all so a step-by-step thing would be good.

Thanks.

2. Originally Posted by GAdams
Hi

Could somebody explain how I would go about doing the attached problem. I don't really get it at all so a step-by-step thing would be good.

Thanks.
In this notation, the nth term of the sequence is identified by the subscript attached to it. So, the first term is $\displaystyle a_1$, the second is $\displaystyle a_2$, etc. you have to find $\displaystyle a_2$ through $\displaystyle a_5$. here, we use the recursive formula given. the formula says:

$\displaystyle a_1 = 1, \mbox { } a_{n + 1} = a_n^2 - 1$

we have to replace the $\displaystyle n$'s in the formula with the appropriate $\displaystyle n$ and evaluate.

so to find $\displaystyle a_2$, we notice that that is $\displaystyle a_{1 + 1}$, so we can take $\displaystyle n = 1$ to find $\displaystyle a_2$

plugging in $\displaystyle n=1$ into the formula, we find that:

$\displaystyle a_2 = a_1^2 - 1$

now evaluate that. the rest are similar

3. I get:

1, 0, -1, 0, -1

4. Originally Posted by GAdams
I get:

1, 0, -1, 0, -1
correct

Give the first five terms.

. . $\displaystyle a_1 = 1\qquad a_{n+1} \:=\:a_n^2-1\qquad(n \,=\,1,2,3,\cdots)$
You have to read what it says . . .

$\displaystyle a_1\,=\,1$
. . It says: the first term is 1.

$\displaystyle a_{n+1} \:=\:a_n^2-1$
. . It says: Each term is the square of the preceding term, minus 1.

So we have:

$\displaystyle \begin{array}{ccccccccc}a_1 & = & 1 \\ a_2 & = & a_1^2 - 1 & = & 1^2-1 & \Rightarrow & a_2 & = & 0\\ a_3 & = & a_2^2-1 & = & 0^2-1 & \Rightarrow & a_3 & = & \text{-}1 \\ a_4 & = & a_3^2-1 & = & (\text{-}1)^2-1 & \Rightarrow & a_4 & = & 0 \\ a_5 & = & a_4^2-1 & = & 0^2-1 & \Rightarrow & a_5 & = & \text{-}1 \end{array}$