Answer
Let $ P (n)$ be the proposition that:
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{n}=\frac{(1-(-7)^{n+1})}{4}
$$
BASIS STEP: P(0) is true, because
by substituting $n=0 $ in the proposition $ P (n)$ , we obtain that $P (0)$ is the statement
$$
2= \frac{(1-(-7)^{0+1})}{4}=\frac{(1-(-7))}{4}=\frac{(8)}{4}=2
$$
(The left-hand side of this equation is 2 because 2 is the sum of the first a nonnegative integer. The right-hand side is found by substituting 0 for $n$ in $\frac{(1-(-7)^{n+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:
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k}=\frac{(1-(-7)^{k+1})}{4}
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k+1}=\frac{(1-(-7)^{k+2})}{4}
$$
is also true. When we add $2\cdot (-7)^{k+1}$ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k+1}= \\
\quad[\text { by the inductive hypothesis }] \\
=\frac{(1-(-7)^{k+1})}{4}+2\cdot (-7)^{k+1}\\
=\frac{1}{4}(1- (-7)^{k+1} +8 \cdot (-7)^{k+1})\\
=\frac{1}{4}( 1- (-7)^{k+2})\quad\quad\quad\quad\quad\quad
\end{split}
$$
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
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{n}=\frac{(1-(-7)^{n+1})}{4}
$$
for all positive integers $n$.
Work Step by Step
Let $ P (n)$ be the proposition that:
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{n}=\frac{(1-(-7)^{n+1})}{4}
$$
BASIS STEP: P(0) is true, because
by substituting $n=0 $ in the proposition $ P (n)$ , we obtain that $P (0)$ is the statement
$$
2= \frac{(1-(-7)^{0+1})}{4}=\frac{(1-(-7))}{4}=\frac{(8)}{4}=2
$$
(The left-hand side of this equation is 2 because 2 is the sum of the first a nonnegative integer. The right-hand side is found by substituting 0 for $n$ in $\frac{(1-(-7)^{n+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:
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k}=\frac{(1-(-7)^{k+1})}{4}
$$
Under this assumption, it must be shown that $P(k + 1)$ is true, namely, that
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k+1}=\frac{(1-(-7)^{k+2})}{4}
$$
is also true. When we add $2\cdot (-7)^{k+1}$ to both sides of the equation in $P(k)$, we obtain
$$
\begin{split}
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{k+1}= \\
\quad[\text { by the inductive hypothesis }] \\
=\frac{(1-(-7)^{k+1})}{4}+2\cdot (-7)^{k+1}\\
=\frac{1}{4}(1- (-7)^{k+1} +8 \cdot (-7)^{k+1})\\
=\frac{1}{4}( 1- (-7)^{k+2})\quad\quad\quad\quad\quad\quad
\end{split}
$$
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
$$
2 -2 \cdot 7+ 2 \cdot 7^{2}-2 \cdot 7^{3}\cdots+2\cdot (-7)^{n}=\frac{(1-(-7)^{n+1})}{4}
$$
for all positive integers $n$.
