I'm stuck and I was wondering if anyone could point me in the right direction
Oh, Im so sorry! I forgot to state what I'm to do with it. It asks me to find a explicit formula for the recursion
$$\begin{align*} a_1&= 3\\ a_n&= 5a_{n-1} + 12 \end{align*}$$
So i found
$$\begin{align*} a_1&=3\\ a_2&=27\\ a_3&= 147\\ a_4&= 747 \end{align*}$$
So far I'm stuck. I can try writing the summation in a different way but I just cannot figure it out
Since it’s a first-order recurrence, you can try the very elementary technique of ‘unwinding’ it:
$$\begin{align*} a_n&=5a_{n-1}+12\\ &=5(5a_{n-2}+12)+12\\ &=5^2a_{n-2}+5\cdot12+12\\ &=5^2(5a_{n-3}+12)+5\cdot12+12\\ &=5^3a_{n-3}+5^2\cdot12+5\cdot12+12\\ &\;\vdots\\ &=5^ka_{n-k}+12(5^{k-1}+5^{k-2}+\ldots+5+1)\tag{1}\\ &\;\vdots\\ &=5^{n-1}a_{n-(n-1)}+12\sum_{k=0}^{n-2}5^k\\ &=5^{n-1}a_1+12\sum_{k=0}^{n-2}5^k\\ &=3\cdot5^{n-1}+12\sum_{k=0}^{n-2}5^k\;. \end{align*}$$
You should know how to express that last summation in closed form, and when you’ve done that, you’ll have a closed form for $a_n$. Of course we guessed at a pattern in line $(1)$, so the closed form that you get should be considered a conjecture, but once you have it, it’s easy enough to prove by induction that it’s correct.
This answer illustrates another, somewhat neater technique for solving exactly this kind of recurrence.
try for characteristic equation : $$a_n=5a_{n-1}+12\\r=5\\a_n=c_1r^n+c_2\\a_n=c_1(5)^n+c_2\\$$apply now $a_1=3,a_2=27$ to find $c_1,c_2$ $$a_1=3=c_15+c_2\\a_2=27=c_125+c_2\\a_n=\frac{6}{5}(5)^n-3\\a_n=6(5)^{n-1}-3$$
If substitute $a_{n-2}$, $a_{n-3}$, and so on, we have: $$a_n=5^k \cdot a_{n-k} + 12 \cdot \sum_{i=0}^{k-1}5^i$$ $$a_n=5^k \cdot a_{n-k} + 12 \cdot {{5^k-1} \over {5-1}}$$ $$a_n=5^k \cdot a_{n-k} + 3 \cdot {(5^k-1)}$$ Let: $k=n-1$, then $n-k=1$ $$a_n=5^{n-1} \cdot a_{1} + 3 \cdot (5^{n-1}-1)$$
