$$3+3 \cdot 5+3 \cdot 5^2+ \cdots +3 \cdot 5^n =\frac{3 \cdot (5^{n+1} -1)}{4}$$
basic step: n=1 $\quad 3+3 \cdot 5 = 3 \cdot (5^{1+1} -1)/4 \iff 18=18 \;$, true
assume: $\quad3+3\cdot 5+3\cdot 5^2+...3\cdot 5^k =3\cdot (5^{k+1} -1)/4$
then:
$3\cdot (5^{n+1} -1)/4 +3 \cdot 5^{n+1} = 3\cdot (5^{n+2} -1)/4$
$3\cdot (5^{n+1} +4\cdot 5^{n+1} -1)/4 = 3\cdot (5^{n+2} -1)/4.$
Could someone explain how the left side can be done to the same as right side with all the steps? It's part of an inductive proof but don't know how to finish it.
You were one small step away. All that's left to do in the last expression is note that $$5^{n+1} + 4 \cdot 5^{n+1} = (1 + 4)\cdot 5^{n+1} = 5 \cdot 5^{n+1} = 5^{n+2}$$
which then gives the sought equality:
$$\frac{3\cdot (5^{n+2} -1)}{4} = \frac{3\cdot (5^{n+2} -1)}{4}$$
A few more notes...
You could have started the induction with $n=0$ as the base case: $3 \cdot 5^0 = 3\cdot(5^1-1)/4$.
You use $k$ in the "assume" part, but $n$ in the induction step.
The left hand side is just 3 times the sum of a geometric progression, so without induction:
$$3\cdot(1 + 5 + \cdots + 5^n) = 3\cdot \frac{5^{n+1}-1}{5-1}$$