# How to find an explicit formula for sequence defined by recurrence relation?

I have gone through the recurrence relation concept.

I have got to know two cases i.e.,

$$1)$$ $$a_n = C^n a_{n-1}$$

$$2)$$ $$F_n = F_{n-1}-F_{n-2}$$

But I got a question that Find the explicit formula for the sequence defined by the recurrence relation

$$a_n = 5\cdot a_{n-1} +3$$ with initial condition $$a_1 = 3$$

I have tried to solve.But i could not match the above the recurrance relation to any of the First order or second order recurrence relation.

Can anyone help me to solve this.Please.

## 2 Answers

For your case, the standard approach is:

$1.$ Find a solution for $a_n=5a_{n-1}$ and,

$2.$ Find a particular solution for $a_n=5\cdot a_{n-1}+3$.

For the case $(1)$ we have geometric sequence which has a solution $h_n=a\cdot 5^n$

For the case $(2)$, the particular solution would be something related to last term $3$. That suggest us a constant solution $p_n=k$. If we replace that constant in the original recurrence we get:

$$k=5k+3 \Rightarrow k=-3/4$$

The general solution will be:

$$a_n=h_n+p_n=a\cdot 5^n-3/4$$

And using that

$$a_1=3= a\cdot 5^1-3/4 \Rightarrow a=3/4$$

and then

$$a_n=\frac{3}{4}\left(5^n-1\right)$$

• I was struck at $p_n =k$. Can you say that how -3/4 came. – Nani Bhavani Dec 12 '16 at 17:42
• Check again. Is that clear? You have to replace at the original equation. – Arnaldo Dec 12 '16 at 17:44
• Yea I got it now..Thank You... :) – Nani Bhavani Dec 12 '16 at 17:45

Often you can offset the sequence to make it simpler. Define $b_n=a_n+c$ for some constant $c$. Then $b_n-c=5(b_{n-1}-c)+3=5b_{n-1}-5c+3$. If you choose $c=\frac 35$ you get a form you know.