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Question

The value of $1^3 - 2^3 + 3^3 - ... + 15^3$ is:

The correct answer is
1856

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:

  1. Express S using sum formulas:

    $S = \sum_{k=1}^{15} k^3 - 2 \times (2^3 + 4^3 + \dots + 14^3)$

    $S = \sum_{k=1}^{15} k^3 - 2 \sum_{j=1}^{7} (2j)^3$

    $S = \sum_{k=1}^{15} k^3 - 2 \sum_{j=1}^{7} 8j^3$

    $S = \sum_{k=1}^{15} k^3 - 16 \sum_{j=1}^{7} j^3$

  2. Calculate the sum of the first 15 cubes:

    $\sum_{k=1}^{15} k^3 = \left(\frac{15(15+1)}{2}\right)^2 = \left(\frac{15 \times 16}{2}\right)^2 = (15 \times 8)^2 = 120^2 = 14400$

  3. Calculate the sum of the first 7 cubes:

    $\sum_{j=1}^{7} j^3 = \left(\frac{7(7+1)}{2}\right)^2 = \left(\frac{7 \times 8}{2}\right)^2 = (7 \times 4)^2 = 28^2 = 784$

  4. 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.

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Similar Questions

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Important Questions from Algebra

  1. Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.

  2. The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
  3. The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
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  5. Let $S = \{(m, n) : m, n \in \{1, 2, 3, \dots, 50\}\}$. If the number of elements $(m, n)$ in S such that $6^m + 9^n$ is a multiple of 5 is p and the number of elements $(m, n)$ in S such that $m + n$ is a square of a prime number is q, then $p + q$ is equal to ____________.
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