$S_0=75$ and $S_{n+1} = \frac13S_n - 6,\quad S_n = ?$. $S_0=75$ and $$S_{n+1} = \frac13S_n - 6$$
How do I solve this relation and create a formula for $S_n$?
So far I got, $S_0 = 75$, $S_1 = 19$, $S_2 = \dfrac13$, $S_3 = -\dfrac{53}9$
How would I go about attacking this? I see no pattern or am I trying to find a formula in a wrong manner?
 A: Hint : look at $T_n=S_n-(-9)$.
Why the $-9$ ? Because $-9$ is the only  constant sequence satisfying the recurrence relation.
A: Break into a homogeneous solution and a particular solution.  The homogeneous solution is simply $A (1/3)^n$.  The particular solution is a constant determined from the equation:
$$B = \frac13 B -9 \implies B = -9$$
So the solution is $A (1/3)^n -9$, and at $n=0$, it is $75$.  Thus $A=84$, and
$$S_n= 84 \left ( \frac13 \right )^n -9$$
A: There are a couple of very elementary ways to attack this problem. You can try to ‘unwrap’ the recurrence:
$$\begin{align*}
S_n&=\frac13S_{n-1}-6\\\\
&=\frac13\left(\frac13S_{n-2}-6\right)-6\\\\
&=\left(\frac13\right)^2S_{n-2}-\frac13\cdot6-6\\\\
&=\left(\frac13\right)^2\left(\frac13S_{n-3}-6\right)-\frac13\cdot6-6\\\\
&=\left(\frac13\right)^3S_{n-3}-\left(\frac13\right)^2\cdot6-\frac13\cdot6-6\\\\
&\;\vdots\\\\
&=\left(\frac13\right)^kS_{n-k}-\left(\frac13\right)^{k-1}\cdot6-\ldots-\left(\frac13\right)^2\cdot6-\frac13\cdot6-6\\\\
&\;\vdots\\\\
&=\left(\frac13\right)^nS_0-\left(\frac13\right)^{n-1}-\ldots-\left(\frac13\right)^2\cdot6-\frac13\cdot6-6\\\\
&=75\left(\frac13\right)^n-6\sum_{k=0}^{n-1}\left(\frac13\right)^k\;,
\end{align*}$$
and the summation in the last line can easily be eliminated if you know the formula for the sum of a finite geometric series. This formula should then be proved by induction on $n$, since at step $k$ we assumed that we’d correctly identified the pattern.
A slicker approach is to make a substitution in order to get a simpler recurrence. Let $T_n=S_n+d$ for some as yet undetermined constant $d$. Then $S_n=T_n-d$, so the recurrence $S_{n+1}=\frac13S_n-6$ becomes
$$T_{n+1}-d=\frac13(T_n-d)-6$$
or, after simplification, $$T_{n+1}=\frac13T_n+\frac23d-6\;.$$
If we set $d=9$, this becomes simply $T_{n+1}=\frac13T_n$, and it’s easy to see that this has the closed form solution $$T_n=\left(\frac13\right)^nT_0\;.$$ Now substitute the actual value of $T_0$, which you can get from $S_0$ and $d$, and reverse the substitution to get a closed form for $S_n$.
