Find the infinite sum of a sequence Define a sequence $a_n$ such that $$a_{n+1}=3a_n+1$$ and $a_1=3$ for $n=1,2,\ldots$. Find the sum $$\sum_{n=1} ^\infty \frac{a_n}{5^n}$$ I am unable to find a general expression for $a_n$. Thanks.
 A: Let $x=\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}$,since
$$\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\dfrac{a_{n}}{5^n}+\dfrac{1}{5^{n+1}}$$
so
$$\sum_{n=1}^{\infty}\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}+\sum_{n=1}^{\infty}\dfrac{1}{5^{n+1}}$$
then we have
$$x-\dfrac{a_{1}}{5}=\dfrac{3}{5}x+\dfrac{\frac{1}{25}}{1-\frac{1}{5}}\Longrightarrow x=\dfrac{3}{8}$$
A: Since we have
$$a_{n+1}+\frac 12=3\left(a_n+\frac 12\right)$$
we have
$$a_n+\frac 12=3^{n-1}\left(a_1+\frac 12\right),$$
i.e.
$$a_n=\frac 72\cdot 3^{n-1}-\frac 12.$$
Now note that we have
$$\sum_{n=1}^{\infty}\frac{a_n}{5^n}=\frac{7}{10}\sum_{n=1}^{\infty}\left(\frac 35\right)^{n-1}-\frac 12\sum_{n=1}^{\infty}\left(\frac 15\right)^n$$
A: $a_{n+1}$ can also be written as a function $f(x+1)=3f(x)+1$.
This has infinitely many solutions the solutions are of the form of,
$\frac{1}{6} [(2c+3)3^x-3]$
This has infinitely many solutions but you gave us an initial condition that will make me be able to calculate the exact function. It is $f(1)=3$.
We must solve:
$\frac{1}{6} [3(2c+3)-3]=3$
Which is equivalent to,
$[3(2c+3)-3]=18$ $=>$
$2c+3=7$ $=>$
$c=2$.
Therefore,
$a_n=\frac{1}{6} [7(3^x)-3]$.
Can you take it from here?
P.S: You may find it more useful to express $a_n$ as $\frac{7}{2} 3^{n-1}-\frac{1}{2}$
A: Hint
Add $\frac 12$ to each side $$a_{n+1}+\frac 12=3a_n+\frac 32=3(a_n+\frac 12)$$ Now, define $b_n=a_n+\frac 12$.
I am sure that you can take from here.
A: HINT:
Let $a_m=b_m+c\implies b_{n+1}+c=3(b_n+c)+1\iff b_{n+1}=b_n+2c+1$ 
Set $2c+1=0$ to get
$$a_{n+1}+\dfrac12=3^1\left[a_n+\dfrac12\right]=\cdots=3^r\left[a_{n-r+1}+\dfrac12\right]=3^n\left[a_1+\dfrac12\right]$$
