Sequences: a(1) = 2, a(n) = a(n-1) - 2

[jshaziza]

The sequence is given by:

a(1) = 2
a(n) = a(n-1) - 2

I put perenthesis around n and n-1 to signify that they are denoted from a.

What threw me off on this problem was the a(n-1) -2 . I know I have to incorporate a(1) = 2 into this formula somehow, but I don't know how. Any help would be great. Thx.

 

[Mrspi]

Re: Sequences

a(1)=2
a(n)=a(n-1)-2
I put perenthesis around n and n-1 to signify that they are denoted from a.
What threw me off on this problem was the a(n-1)-2. I know I have to incorporate a(1)= 2 into this formula somehow but I don't know how. Any help would be great. Thx.

You know that a(1) = 2

Ok...

now, you are also told that a(n) = a(n - 1) - 2

So, if you're looking for a(2), you'd have

a(2) = a(2 - 1) - 2
or,
a(2) = a(1) - 2

BUT, you know that a(1) = 2, right? So,

a(2) = 2 - 2
a(2) = 0

If you are asked to find a(3), then use the definition

a(n) = a(n - 1) - 2

a(3) = a(2) - 2

We've just established that a(2) = 0. So,
a(3) = 0 - 2
a(3) = -2

If you need further help, please tell us what you've tried, and where you are "stuck" with this.

 

[soroban]

Re: Sequences

Hello, jshaziza!

Just read what they gave you . . .

\(\displaystyle a_1\:=\:2\)
. . The first term is 2.

\(\displaystyle a_n\:=\:a_{n-1}\,-\,2\)
. . Each term is the preceding term, minus 2.

\(\displaystyle \begin{array}{cccc}\text{The }1^{st}\text{ term is: }&2 \\ \\ \\ \text{The }2^{nd}\text{ term is: } & 2\,-\,2 & = & 0 \\ \\ \\ \text{The }3^{rd}\text{ term is: } & 0\,-\,2 & = & -2\\ \\ \\ \text{The }4^{th}\text{ term is: } & -2\,-\,2 & = & -4 \\ \\ \\ \vdots & & \vdots\end{array}\)