# Thread: Solutions to recurrence relations

1. ## Solutions to recurrence relations

Could anyone point me in the right direction with these two questions, I am lost on where to start.

An=-An-1, a0 = 5 (Note: n-1 is a subscript)
An=An-1 + 3, a0 = 1

2. Hello, vexiked1

You haven't told us what we're supposed to do.
But I'll assume we want the formula for the $n^{th}$ term.

$(1)\;\;A_n \:=\:-A_{n-1},\;\; A_0 = 5$

It says: each term is the negative of the preceding term.

So the sequence is: . $5,\:\text{-}5,\;5,\:\text{-}5,\;5,\:\text{-}5,\;\hdots$

The general term is: . $A_n \:=\:(\text{-}1)^{n-1}\cdot 5$

$(2)\;\;A_n\:=\:A_{n-1} + 3,\;\;A_0 = 1$

It says: Each term is three more than the preceding term.

So the sequence is: . $1,\:4,\:7,\:10,\:13,\:\hdots$

The general term is: . $A_n \;=\;3n-2$

3. My apologies.... I am to find the solution to the recurrence relations with the given initial conditions. So does this still apply? So the basic approach is to find the sequence first then?

I am also having trouble with:

An=2An-1 -3, a0 = -1
An=(n+1)An-1, a0=2
An=2nAn-1, a0=3
An=-An-1 + n - 1, a0 = 7