Find $a_n$, given $a_1=1$ and $a_{n+1} =a_n/4+3/4$ The sequence is defined of rational numbers $a_1,a_2,a_3,...$. Let $a_1=1$ and $a_{n+1}=a_n/4+3/4$. What are the numbers $a_1,a_2,a_3,...$ ?
What I've done:
$a_1=1$
for $n=1$ we have $a_{n+1}=a_2=a_1/4+3/4=1$
for $n=2$ we have $a_{n+1}=a_3=a_2/4+3/4=1$
Therefore the sequence consists of the numbers $1,1,1,...$
I have two questions.
1) Is it correct or am I misunderstanding something?
2) How would I go about finding $a_n$?
 A: Claim: for every $n \in \mathbb{N}$; we have: $a_n=1$ .
Proof by induction: suppose that the assertion holds for $k=n$; 
i.e. $a_n=1$ ; then one can see that 
$$a_{n+1}=\dfrac{a_n}{4}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1;$$ 
so the assertion also holds for $k=n+1$. 

Let $A \in \mathbb{R}$ be arbitrery, and let $a_1=A$; 
Claim: for every $n \in \mathbb{N}$; we have: $a_n=\dfrac{A-1}{4^{n-1}}+1$ .
Proof by induction: suppose that the assertion holds for $k=n$; 
i.e. $a_n=\dfrac{A-1}{4^{n-1}}+1$ ; then one can see that 
$$a_{n+1}=\dfrac{a_n}{4}+\dfrac{3}{4}=\dfrac{\dfrac{A-1}{4^{n-1}}+1}{4}+\dfrac{3}{4}= 
\dfrac{A-1}{4^{n}}+\dfrac{1}{4}+\dfrac{3}{4} 
= 
\dfrac{A-1}{4^{n}}+1 
; 
$$ 
so the assertion also holds for $k=n+1$. 

Second proof: 
Let's define $b_n=a_n-1$, so we get the second sequence as follows:  
$$ 
b_0=A-1 
\ \ \ \ 
\text{and} 
\ \ \ \ 
b_n=\dfrac{b_n}{4} 
; 
$$ 
one can see easilly that $b_n=\dfrac{A-1}{4^{n-1}}$, 
so we can conclude $a_n=\dfrac{A-1}{4^{n-1}}+1$ . 
A: To find $a_n$, express the given recurrence equation as:
$$a_{n+1}-1=\frac{1}{4}(a_n-1), a_1=1$$
Denoting $b_n=a_n-1, b_1=a_1-1=0$ results in:
$$b_{n+1}=\frac14b_n \Rightarrow b_n=b_1\cdot \left(\frac14\right)^{n-1}=0.$$
Hence:
$$0=b_n=a_n-1 \Rightarrow a_n=1.$$
A: You are right: we have $a_n=1$ for all $n$. You can prove this result by induction.
A: I have previously developed a general solution for all problems of the type $a_n=Aa_{n-1}+B$, as shown here. This solution is a bit more complicated than the others preferred here, but again, it applies to any such problem and the work is already done. The solution was shown to be
$$a_n=\frac{[(A-1)a_0+B]A^n-B}{A-1}$$
and you can readily show that $a_n=1 ~\forall ~n$ in the present case.
