Sequences: If a(n) = [a(n-1) - 2]^2 and a(1) = 1, then....

[TheNextOne]

Hi, I need help with this question.

If a sequence is defined by the rule a(n) = [a(n-1)-2]^2 what is a(4) if a(1) is 1?

The (N) is suppose to be a subscript. I have the answer as 4 but dont understand it.

 

[pka]

Re: Sequences

If a sequence is defined by the rule a(n) = [a(n-1)-2]^2 what is a(4) if a(1) is 1?

a<SUB>2</SUB>=[a<SUB>1</SUB>-2]<SUP>2</SUP>=[1-2]<SUP>2</SUP>=1.
a<SUB>3</SUB>=[a<SUB>2</SUB>-2]<SUP>2</SUP>=[1-2]<SUP>2</SUP>=1.
a<SUB>4</SUB>=?

 

[TheNextOne]

sorry, its suppose to be -3 instead of -2 at the end of the equation.

 

[stapel]

sorry, its suppose to be -3 instead of -2 at the end of the equation.

That's fine. Replace the correct number, and then follow the displayed process to find your answer.

Eliz.

 

[soroban]

Re: Sequences

Hello, TheNextOne!

If a sequence is defined by the rule: \(\displaystyle a_n \:= \:[a_{n-1}\,-\,3]^2\)
\(\displaystyle \;\;\)what is \(\displaystyle a_4\) if \(\displaystyle a_1\,=\,1\)?

This is a recurrence. \(\displaystyle \;\)You must be able to "read" that formula.
\(\displaystyle \;\;\)Use baby-talk if necessary . . .


We are given: \(\displaystyle \:a_n\;=\;(a_{n-1}\,-\,3)^2\)

It says: To get the \(\displaystyle n^{th}\) term, take the preceding term \(\displaystyle (a_{n-1})\), subtract 3, then square.

We know that: \(\displaystyle a_1\,=\,1\)

To find \(\displaystyle a_2\), take \(\displaystyle a_1\). subtract 3, and square.
\(\displaystyle \;\;a_2\:=\1\,-\, 3)^2\:=\-2)^2\:=\:4\)

To find \(\displaystyle a_3\), repeat the process with \(\displaystyle a_2.\)
\(\displaystyle \;\;a_3\:=\4\,-\,3)^2\:=\:1^2\:=\:1\)

To find \(\displaystyle a_4\), repeat the process with \(\displaystyle a_3.\)
\(\displaystyle \;\;a_4\:=\1\,-\,3)^2\:=\-2)^2\:=\:4\;\;\) . . . There!


If there is a part (b) to this problem, I bet it says something like: \(\displaystyle \text{Find }a_{37}\)
\(\displaystyle \;\;\)Can you find it without cranking through thrity-six steps?