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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.

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2 Answers 2

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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

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

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  • $\begingroup$ I was struck at $ p_n =k $. Can you say that how -3/4 came. $\endgroup$ Commented Dec 12, 2016 at 17:42
  • $\begingroup$ Check again. Is that clear? You have to replace at the original equation. $\endgroup$
    – Arnaldo
    Commented Dec 12, 2016 at 17:44
  • $\begingroup$ Yea I got it now..Thank You... :) $\endgroup$ Commented Dec 12, 2016 at 17:45
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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.

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