Answer
See the proof below.
Work Step by Step
Consider the sequence $2^1,2^2,2^3,2^4,\ldots$. For each positive integer let
$$S_n=2^1+2^2+2^3+\ldots+2^n.$$
Let's calculate a few values of $S_n$ so we can look for pattern and then propose a formula for $S_n$.
Observe that
$$S_1=2^1=2,$$
$$S_2=2^1+2^2=2+4=6,$$
$$S_3=2^1+2^2+2^3=2+4+8=14,$$
and
$$S_4=2^1+2^2+2^3+2^4=2+4+8+16=30.$$
We can see that
$S_1+2=2+2=4=2^2$, which implies that $S_1=2^2-2=2^{1+1}-2$,
$S_2+2=6+2=8=2^3$, which implies that $S_2=2^3-2=2^{2+1}-2$,
$S_3+2=14+2=16=2^4$, which implies that $S_3=2^4-2=2^{3+1}-2$,
and
$S_4+2=30+2=32=2^5$, which implies that $S_4=2^5-2=2^{4+1}-2$.
The previous pattern leads us to suspect that for any positive integer it is true that
$$S_n=2^{n+1}-2.$$
We will use mathematical induction to show that our guess is correct.
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In order to show that the formula holds we must show that it is true for $n=1$ and that if its is true for $n=k$ then, it is true for $n=k+1$.
$\textbf{1.}$ The formula holds for $n=1$ since
$$S_1=2^1=2=2^{2+1}-2.$$
$\textbf{2.}$ Suppose the formula holds for $n=k$, that is, suppose that
$$S_k=2^1+2^2+2^3+\ldots+2^k=2^{k+1}-2.$$
Then, adding $2^{k+1}$ to the equality $2^1+2^2+2^3+\ldots+2^k=2^{k+1}-2$
gives us that
$$S_{k+1}=2^1+2^2+2^3+\ldots+2^k+2^{k+1}=2^{k+1}-2+2^{k+1}.$$
Now, observe that
$$2^{k+1}-2+2^{k+1}=2\left(2^{k+1}\right)-2=2^1\left(2^{k+1}\right)-2=2^{\left(k+1\right)+1}-2.$$
Thus,
$$S_{k+1}=2^{\left(k+1\right)+1}-2$$
and therefore, the formula holds for $n=k+1$ whenever it is true for $n=k$.
Hence, the mathematical induction principle guarantees that the formula is true for every positive integer $n$.
