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. 
 A: 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.
A: I imagine the post on how to write a clear induction proof could be of great service to you. Bob's answer highlights the key points, but I thought I would provide another answer to possibly increase clarity. 
You have completed the base case and that's the first part. Great. Now, fix some integer $k\geq 0$ and assume that the statement
$$
S(k) : \color{green}{\sum_{i=0}^k3\cdot5^i=\frac{3(5^{k+1}-1)}{4}}
$$
holds (this is the inductive hypothesis). To be shown is that
$$
S(k+1) : \color{blue}{\sum_{i=0}^{k+1}3\cdot5^i=\frac{3(5^{k+2}-1)}{4}}
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\color{blue}{\sum_{i=0}^{k+1}3\cdot5^i}&= \color{green}{\sum_{i=0}^k3\cdot5^i}+3\cdot5^{k+1}\tag{by definition of $\Sigma$}\\[1em]
&= \color{green}{\frac{3(5^{k+1}-1)}{4}}+3\cdot5^{k+1}\tag{by inductive hypothesis}\\[1em]
&= \frac{3(5^{k+1}-1)+4\cdot3\cdot5^{k+1}}{4}\tag{common denominator}\\[1em]
&= \frac{3\cdot5^{k+1}-3+12\cdot5^{k+1}}{4}\tag{simplify}\\[1em]
&= \frac{15\cdot5^{k+1}-3}{4}\tag{simplify}\\[1em]
&= \color{blue}{\frac{3(5^{k+2}-1)}{4}},\tag{factor / simplify}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step. Thus, the statement $S(n)$ is true for all integers $n\geq0$. $\blacksquare$
A: 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, $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{n}}=\dfrac{3(5^{n+1}-1)}{4}$; $n\geq{0}$.
