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. 

Let $S(n)$ be the statement: $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{n}}=\dfrac{3(5^{n+1}-1)}{4}$; $n\geq{0}$
Basis step: $S(0)$:
LHS: $3\times{5^{0}}=3\times{1}$
$\hspace{23 mm}=3$
RHS: $\dfrac{3(5^{(0)+1}-1)}{4}=\dfrac{3(5^{1}-1)}{4}$
$\hspace{37.5 mm}=\dfrac{3\times{4}}{4}$
$\hspace{37.5 mm}=3$
$\hspace{75 mm}$LHS $=$ RHS (verified.)
Induction step:
Assume $S(k)$ is true, i.e. assume that $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{k}}=\dfrac{3(5^{k+1}-1)}{4}$; $k\geq{0}$
$S(k+1)$: $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{k}}+3\times{5^{k+1}}=\dfrac{3(5^{k+1}-1)}{4}+3\times{5^{k+1}}$
$\hspace{97.5 mm}=\dfrac{3(5^{k+1}-1)+4(3\times{5^{k+1}})}{4}$
$\hspace{97.5 mm}=\dfrac{3\times{5^{k+1}}-3+12\times{5^{k+1}}}{4}$
$\hspace{97.5 mm}=\dfrac{15\times{5^{k+1}}-3}{4}$
$\hspace{97.5 mm}=\dfrac{3(5\times{5^{k+1})}-3}{4}$
$\hspace{97.5 mm}=\dfrac{3(5^{k+2}-1)}{4}$
So, $S(k+1)$ is true whenever $S(k)$ is true.
Therefore, $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{n}}=\dfrac{3(5^{n+1}-1)}{4}$; $n\geq{0}$.
