Solving the Alternating Sum of Cubes: $1^3 - 2^3 + \dots + 15^3$
This problem requires calculating the value of the series $S = 1^3 - 2^3 + 3^3 - 4^3 + \dots + 15^3$. This is an alternating sum of cubes.
Strategy for Calculation
We can express the series $S$ using the standard sum of cubes formula $\sum_{k=1}^{N} k^3 = \left(\frac{N(N+1)}{2}\right)^2$.
Rewrite the series by separating positive and negative terms:
$S = (1^3 + 3^3 + 5^3 + \dots + 15^3) - (2^3 + 4^3 + 6^3 + \dots + 14^3)$
A more efficient method is to express $S$ in terms of the sum of all cubes up to 15 and subtract twice the sum of the even cubes:
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Express S using sum formulas:
$S = \sum_{k=1}^{15} k^3 - 2 \times (2^3 + 4^3 + \dots + 14^3)$
$S = \sum_{k=1}^{15} k^3 - 2 \sum_{j=1}^{7} (2j)^3$
$S = \sum_{k=1}^{15} k^3 - 2 \sum_{j=1}^{7} 8j^3$
$S = \sum_{k=1}^{15} k^3 - 16 \sum_{j=1}^{7} j^3$
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Calculate the sum of the first 15 cubes:
$\sum_{k=1}^{15} k^3 = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = (15 \times 8)^2 = 120^2 = 14400$
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Calculate the sum of the first 7 cubes:
$\sum_{j=1}^{7} j^3 = \left(\frac{7(7+1)}{2}\right)^2 = \left(\frac{7 \times 8}{2}\right)^2 = (7 \times 4)^2 = 28^2 = 784$
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Substitute the sums back into the expression for S:
$S = 14400 - 16 \times 784$
$S = 14400 - 12544$
$S = 1856$
Final Result
The calculated value of the series $1^3 - 2^3 + 3^3 - \dots + 15^3$ is 1856.
Similar Questions
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