0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ I was struck at $ p_n =k $. Can you say that how -3/4 came. $\endgroup$ – Nani Bhavani Dec 12 '16 at 17:42
  • $\begingroup$ Check again. Is that clear? You have to replace at the original equation. $\endgroup$ – Arnaldo Dec 12 '16 at 17:44
  • $\begingroup$ Yea I got it now..Thank You... :) $\endgroup$ – Nani Bhavani Dec 12 '16 at 17:45
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.