Key Concepts and Formulas

• Sum of the First $n$ Cubes: The sum of the cubes of the first $n$ positive integers is given by the formula: $\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + 3^3 + \dots + n^3 = \left( \frac{n(n+1)}{2} \right)^2$
• Algebraic Manipulation of Series: Alternating series can often be simplified by grouping terms and using algebraic identities to transform them into standard series for which summation formulas are known.
• Properties of Exponents: $(ab)^n = a^n b^n$. This is useful for simplifying sums of powers of even numbers.

Step-by-Step Solution

Step 1: Analyze the Given Series and Rewrite It The given series is $1^3 - 2^3 + 3^3 - 4^3 + \dots + 9^3$. This is an alternating series of cubes. Let $S$ be the sum of this series. $S = 1^3 - 2^3 + 3^3 - 4^3 + 5^3 - 6^3 + 7^3 - 8^3 + 9^3$ To effectively use the sum of cubes formula, we can rewrite this series by separating the positive and negative terms: $S = (1^3 + 3^3 + 5^3 + 7^3 + 9^3) - (2^3 + 4^3 + 6^3 + 8^3)$ A common strategy for such alternating series is to express it as the sum of all terms up to the last term, minus twice the sum of the terms that have a negative sign. We can achieve this by adding and subtracting the even-powered terms: $S = (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3) - (2^3 + 4^3 + 6^3 + 8^3) - (2^3 + 4^3 + 6^3 + 8^3)$ $S = \left(\sum_{k=1}^{9} k^3\right) - 2 \left(2^3 + 4^3 + 6^3 + 8^3\right)$ This step is crucial as it allows us to use the standard sum of cubes formula for the first part and a modified version for the second part.

Step 2: Calculate the Sum of All Cubes up to $9^3$ The first part of our rewritten series is the sum of the cubes of the first 9 positive integers, $\sum_{k=1}^{9} k^3$. We use the formula $\left( \frac{n(n+1)}{2} \right)^2$ with $n=9$: $\sum_{k=1}^{9} k^3 = \left( \frac{9(9+1)}{2} \right)^2 = \left( \frac{9 \times 10}{2} \right)^2$ $= \left( \frac{90}{2} \right)^2 = (45)^2$ Calculating $(45)^2$: $45 \times 45 = (40+5)(40+5) = 1600 + 200 + 200 + 25 = 2025$ So, the sum of the first 9 cubes is 2025.

Step 3: Calculate Twice the Sum of the Even Cubes The second part of our rewritten series is $2 \times (2^3 + 4^3 + 6^3 + 8^3)$. We need to evaluate the sum of the even cubes first. Each term is a cube of an even number, which can be written as $(2k)^3$: $2^3 + 4^3 + 6^3 + 8^3 = (2 \times 1)^3 + (2 \times 2)^3 + (2 \times 3)^3 + (2 \times 4)^3$ Using the property of exponents $(ab)^n = a^n b^n$: $= 2^3 \cdot 1^3 + 2^3 \cdot 2^3 + 2^3 \cdot 3^3 + 2^3 \cdot 4^3$ Now, we can factor out $2^3 = 8$: $= 8 (1^3 + 2^3 + 3^3 + 4^3)$ The sum inside the parenthesis is the sum of the cubes of the first 4 positive integers. We apply the sum of cubes formula with $n=4$: $1^3 + 2^3 + 3^3 + 4^3 = \left( \frac{4(4+1)}{2} \right)^2 = \left( \frac{4 \times 5}{2} \right)^2$ $= \left( \frac{20}{2} \right)^2 = (10)^2 = 100$ So, the sum of the even cubes is $8 \times 100 = 800$. Now, we need to calculate twice this sum: $2 \times (800) = 1600$

Step 4: Combine the Results to Find the Final Sum Now we substitute the values calculated in Step 2 and Step 3 back into our rewritten series equation from Step 1: $S = \left(\sum_{k=1}^{9} k^3\right) - 2 \left(2^3 + 4^3 + 6^3 + 8^3\right)$ $S = 2025 - 1600$ $S = 425$

Common Mistakes & Tips

• Incorrectly Applying the Sum Formula: Ensure that when using the formula $\left( \frac{n(n+1)}{2} \right)^2$, you use the correct value of $n$ for each specific sum (e.g., $n=9$ for the sum of the first 9 cubes, and $n=4$ for the sum of the first 4 cubes).
• Algebraic Errors with Even Terms: When dealing with sums of even powers like $2^3 + 4^3 + \dots$, remember to correctly factor out the common base power (in this case, $2^3$) to reduce it to a sum of consecutive integer powers.
• Arithmetic Mistakes: Carefully perform the squaring and multiplication operations. For example, $(45)^2 = 2025$ and $16 \times 100 = 1600$ are key calculations.

Summary

The problem asks for the sum of an alternating series of cubes. The strategy employed is to rewrite the alternating series into a form where the standard sum of cubes formula can be applied. This involves expressing the sum as the sum of all cubes up to $9^3$ minus twice the sum of the even cubes ($2^3, 4^3, 6^3, 8^3$). The sum of the first 9 cubes is calculated using the formula $\left( \frac{n(n+1)}{2} \right)^2$ with $n=9$. The sum of the even cubes is simplified by factoring out $2^3$ and then using the same sum of cubes formula for the remaining sum of consecutive cubes. Finally, these two parts are combined to obtain the result.

The final answer is $\boxed{425}$.