Proving that $3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n = [3(5^{n+1} - 1)] / 4$ whenever $n \geq 0$ 
Use induction to show that $$3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n= \frac{3(5^{n+1} - 1)}{4} $$whenever $n$ is a non-negative integer.

I know I need a base-case where $n = 0$:
$$3 \times 5^0 = \frac{3(5^{0+1} - 1)}{4}\\LHS = 3  = \frac{12}{4} = RHS$$
Next I need to show that this is true for the $n + 1$ (next) term through a proof using induction. This is really where I could use a concrete example of a proof; I have yet to find one that I could really understand. 

You are trying to show that 
$$\sum_{n=0}^{N}5^n=\frac{5^{N+1}-1}{4}$$
We can leave out the factor of $3$ since it just multiplies both sides. The base case is simple, you just have $1=1$. Now assume it is true for $N$. Then we have
$$\sum_{n=0}^{N+1}5^n=\sum_{n=0}^{N}5^n+5^{N+1}=\frac{5^{N+1}-1}{4}+5^{N+1}=\frac{5^{N+1}-1+4\cdot 5^{N+1}}{4}=\frac{5\cdot 5^{N+1}-1}{4}$$
Which gives
$$\sum_{n=0}^{N+1}5^{n}=\frac{5^{(N+1)+1}-1}{4}$$
Which proves that the statement is true for $N+1$. Thus, the statement is true by induction.
