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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.

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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 $$x-\dfrac{a_{1}}{5}=\dfrac{3}{5}x+\dfrac{\frac{1}{25}}{1-\frac{1}{5}}\Longrightarrow x=\dfrac{3}{8}$$

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  • $\begingroup$ Nice, +1. But please see , you did an error. You would get $x=\frac{13}{8}$ $\endgroup$ – user167045 Jun 23 '15 at 14:03
  • $\begingroup$ @You-know-me,Thank you $\endgroup$ – math110 Jun 23 '15 at 14:09
  • $\begingroup$ Could you edit the answer to $\frac{13}{8}$. It is not letting me edit. $\endgroup$ – Akash Gaur Mar 21 '19 at 8:20
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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 $$\sum_{n=1}^{\infty}\frac{a_n}{5^n}=\frac{7}{10}\sum_{n=1}^{\infty}\left(\frac 35\right)^{n-1}-\frac 12\sum_{n=1}^{\infty}\left(\frac 15\right)^n$$

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$a_{n+1}$ 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 $\frac{7}{2} 3^{n-1}-\frac{1}{2}$

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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.

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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

$$a_{n+1}+\dfrac12=3^1\left[a_n+\dfrac12\right]=\cdots=3^r\left[a_{n-r+1}+\dfrac12\right]=3^n\left[a_1+\dfrac12\right]$$

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