Let's solve the problem and show every step:

We are tasked to prove that: $\sum_{k=1}^n (-1)^k k^2 = \frac{(-1)^n n (n+1)}{2}, \quad \text{for } n = 1, 2, 3, \dots.$

Step 1: Analyze the summation formula

The given sum is: $S_n = \sum_{k=1}^n (-1)^k k^2.$

Here, $(-1)^k$ alternates the sign of each term depending on whether $k$ is odd ($(-1)^k = -1$) or even ($(-1)^k = 1$). Each term in the sum is $(-1)^k \cdot k^2$, where $k^2$ is the square of the current index $k$.

We will use induction to prove the formula: $S_n = \frac{(-1)^n n (n+1)}{2}.$

Step 2: Base case ($n = 1$)

For $n = 1$, the summation is: $S_1 = \sum_{k=1}^1 (-1)^k k^2 = (-1)^1 \cdot 1^2 = -1.$

The right-hand side of the formula for $n = 1$ is: $\frac{(-1)^n n (n+1)}{2} = \frac{(-1)^1 \cdot 1 \cdot (1+1)}{2} = \frac{-1 \cdot 1 \cdot 2}{2} = -1.$

Thus, the formula holds for $n = 1$.

Step 3: Inductive hypothesis

Assume that the formula holds for $n = m$, i.e., $S_m = \sum_{k=1}^m (-1)^k k^2 = \frac{(-1)^m m (m+1)}{2}.$

Step 4: Inductive step ($n = m+1$)

We need to prove that the formula holds for $n = m+1$, i.e., that: $S_{m+1} = \sum_{k=1}^{m+1} (-1)^k k^2 = \frac{(-1)^{m+1} (m+1) ((m+1)+1)}{2}.$

From the definition of $S_{m+1}$, we can write: $S_{m+1} = S_m + (-1)^{m+1} (m+1)^2.$

Using the inductive hypothesis $S_m = \frac{(-1)^m m (m+1)}{2}$, substitute $S_m$ into the equation: $S_{m+1} = \frac{(-1)^m m (m+1)}{2} + (-1)^{m+1} (m+1)^2.$

Step 5: Combine terms

Factor out $(m+1)$ from the terms: $S_{m+1} = (m+1) \left[ \frac{(-1)^m m}{2} + (-1)^{m+1} (m+1) \right].$

Simplify the bracketed expression. Rewrite $(-1)^{m+1}$ as $-(-1)^m$: $\frac{(-1)^m m}{2} + (-1)^{m+1} (m+1) = \frac{(-1)^m m}{2} - (-1)^m (m+1).$

Combine the terms: $\frac{(-1)^m m}{2} - (-1)^m (m+1) = \frac{(-1)^m m - 2(-1)^m (m+1)}{2}.$

Distribute $(-1)^m$ in the numerator: $\frac{(-1)^m m - 2(-1)^m (m+1)}{2} = \frac{(-1)^m (m - 2m - 2)}{2}.$

Simplify the numerator: $(-1)^m (m - 2m - 2) = (-1)^m (-m - 2) = (-1)^m (-(m+2)).$

Thus: $\frac{(-1)^m m}{2} + (-1)^{m+1} (m+1) = \frac{(-1)^m (-(m+2))}{2}.$

Step 6: Final simplification

Substitute back into $S_{m+1}$: $S_{m+1} = (m+1) \cdot \frac{(-1)^m (-(m+2))}{2}.$

Simplify: $S_{m+1} = \frac{(-1)^m (m+1)(-(m+2))}{2}.$

Distribute the negative sign: $S_{m+1} = \frac{(-1)^{m+1} (m+1)(m+2)}{2}.$

This matches the formula: $S_{m+1} = \frac{(-1)^{m+1} (m+1)((m+1)+1)}{2}.$

Thus, the formula holds for $n = m+1$.

Step 7: Conclusion

By induction, the formula: $\sum_{k=1}^n (-1)^k k^2 = \frac{(-1)^n n (n+1)}{2}$ is true for all $n = 1, 2, 3, \dots$.

Questions for further exploration:

1. What happens if the summation alternates with $k^3$ instead of $k^2$?
2. How can this formula be derived using properties of finite series?
3. What is the behavior of the series as $n \to \infty$?
4. Can this result be extended to other powers of $k$, such as $k^4$?
5. How does this formula relate to combinatorial or geometric interpretations?

Tip:

When using induction, always verify the base case and ensure your inductive step properly accounts for the next term.