We are asked to prove the following statement by mathematical induction:

$1 - 2 + 2^2 - 2^3 + \cdots + (-1)^n 2^n = \frac{2^{n+1} (-1)^n + 1}{3}$

Step 1: Base Case (n = 1)

For $n = 1$, the left-hand side of the equation becomes:

$1 - 2 = -1$

Now, let's check the right-hand side of the equation when $n = 1$:

$\frac{2^{1+1} (-1)^1 + 1}{3} = \frac{2^2 (-1) + 1}{3} = \frac{4(-1) + 1}{3} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1$

So, for $n = 1$, both sides are equal. The base case holds.

Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer $n = k$. That is, assume:

$1 - 2 + 2^2 - 2^3 + \cdots + (-1)^k 2^k = \frac{2^{k+1} (-1)^k + 1}{3}$

We need to prove that the statement is true for $n = k+1$, i.e.,

$1 - 2 + 2^2 - 2^3 + \cdots + (-1)^{k+1} 2^{k+1} = \frac{2^{(k+1)+1} (-1)^{k+1} + 1}{3}$

Step 3: Prove for $n = k+1$

Starting from the inductive hypothesis, we need to add the next term $(-1)^{k+1} 2^{k+1}$ to both sides:

$\left( 1 - 2 + 2^2 - 2^3 + \cdots + (-1)^k 2^k \right) + (-1)^{k+1} 2^{k+1} = \frac{2^{k+1} (-1)^k + 1}{3} + (-1)^{k+1} 2^{k+1}$

Simplifying the right-hand side:

$\frac{2^{k+1} (-1)^k + 1}{3} + (-1)^{k+1} 2^{k+1} = \frac{2^{k+1} (-1)^k + 1 + 3(-1)^{k+1} 2^{k+1}}{3}$

Factor out $2^{k+1}$ from the terms involving powers of 2:

$= \frac{2^{k+1} \left( (-1)^k + 3(-1)^{k+1} \right) + 1}{3}$

Now, simplify the expression inside the parentheses:

$(-1)^k + 3(-1)^{k+1} = (-1)^k - 3(-1)^k = -2(-1)^k$

Substitute this back into the equation:

$= \frac{2^{k+1} (-2)(-1)^k + 1}{3} = \frac{2^{k+2} (-1)^{k+1} + 1}{3}$

This is exactly the right-hand side of the equation for $n = k+1$.

Conclusion

By the principle of mathematical induction, the given statement is true for all positive integers $n$.

Do you want further details or have any questions?

Here are 5 related questions you might explore:

1. How does mathematical induction work in more complex problems?
2. Can we apply a different proof method for this type of summation problem?
3. How would this proof change if the powers of 2 were replaced by another sequence?
4. What are some variations of alternating series that can be solved by induction?
5. Can this method be generalized to other powers or polynomial forms?

Tip: Always verify your base case carefully before moving on to the inductive step in any proof by induction.