Answer
Let $ P (n)$ be the proposition that:
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{n}=\frac{3(5^{n+1}-1)}{4}
$$
BASIS STEP: P(0) is true, because
by substituting $n=0 $ in the proposition $ P (n)$ , we obtain that $P (0)$ is the statement
$$
3= \frac{3(5^{0+1}-1)}{4}=\frac{3(5-1)}{4}=3
$$
(The left-hand side of this equation is 3 because 3 is the sum of the first a nonnegative integer. The right-hand side is found by substituting 3 for $n$ in $\frac{3(5^{n+1}-1)}{4} $)
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k) holds for an arbitrary
a nonnegative integer $k$. That is, we assume that:
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{k}=\frac{3(5^{k+1}-1)}{4}
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{k+1}=\frac{3(5^{k+2}-1)}{4}
$$
is also true. When we add $3 \cdot 5^{k+1}$ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots +3 \cdot 5^{k+1}= \\
\quad[\text { by the inductive hypothesis }] \\
=\frac{3}{4}(5^{k+1}-1 +4 \cdot 5^{k+1})\\
=\frac{3}{4}(5 \cdot 5^{k+1}-1)\\
=\frac{3}{4}( 5^{k+2}-1)
\end{split}
$$
We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.
That is, we have proven that
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{n}=\frac{3(5^{n+1}-1)}{4}
$$
for all non-negative integers $n$.
Work Step by Step
Let $ P (n)$ be the proposition that:
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{n}=\frac{3(5^{n+1}-1)}{4}
$$
BASIS STEP: P(0) is true, because
by substituting $n=0 $ in the proposition $ P (n)$ , we obtain that $P (0)$ is the statement
$$
3= \frac{3(5^{0+1}-1)}{4}=\frac{3(5-1)}{4}=3
$$
(The left-hand side of this equation is 3 because 3 is the sum of the first a nonnegative integer. The right-hand side is found by substituting 3 for $n$ in $\frac{3(5^{n+1}-1)}{4} $)
INDUCTIVE STEP:
For the inductive hypothesis we assume that P(k) holds for an arbitrary
a nonnegative integer $k$. That is, we assume that:
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{k}=\frac{3(5^{k+1}-1)}{4}
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{k+1}=\frac{3(5^{k+2}-1)}{4}
$$
is also true. When we add $3 \cdot 5^{k+1}$ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots +3 \cdot 5^{k+1}= \\
\quad[\text { by the inductive hypothesis }] \\
=\frac{3}{4}(5^{k+1}-1 +4 \cdot 5^{k+1})\\
=\frac{3}{4}(5 \cdot 5^{k+1}-1)\\
=\frac{3}{4}( 5^{k+2}-1)
\end{split}
$$
We have completed the basis step and the inductive step, so by mathematical induction we know that $P(n)$ is true for all positive integers $n$.
That is, we have proven that
$$
3 +3 \cdot 5+ 3 \cdot 5^{2}+3 \cdot 5^{3}\cdots+3 \cdot 5^{n}=\frac{3(5^{n+1}-1)}{4}
$$
for all positive integers $n$.
