Question

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

• A. 1706
• B. 1856
• C. 2028
• D. 2256

Hint:

Write the series as $(1^3 + 2^3 + ... + 15^3) - 2(2^3 + 4^3 + ... + 14^3)$ and use the formula $\sum r^3 = [n(n+1)/2]^2$.

Answer 1

The Correct Option is B

We can use the general formula for the alternating sum of cubes: $\sum_{k=1}^n (-1)^{k-1} k^3$.
If $n$ is odd, the formula is $S_n = \frac{(n+1)^2(2n-1)}{4}$.

In our case, $n = 15$, which is an odd number.
Substituting $n = 15$ into the formula:
$S_{15} = \frac{(15+1)^2(2 \times 15 - 1)}{4}$
$S_{15} = \frac{16^2 imes 29}{4}$
$S_{15} = \frac{256 imes 29}{4}$
$S_{15} = 64 imes 29$

Let's calculate $64 \times 29$:
$64 \times 30 = 1920$
$64 \times 1 = 64$
$1920 - 64 = 1856$.

Alternatively, grouping terms:
$1^3 + (3^3 - 2^3) + (5^3 - 4^3) + ... + (15^3 - 14^3)$
Using $a^3 - b^3 = (a-b)(a^2+ab+b^2)$, where $a-b=1$:
$S = 1 + \sum_{k=1}^7 ( (2k+1)^2 + (2k+1)(2k) + (2k)^2 )$
$S = 1 + \sum_{k=1}^7 ( 4k^2+4k+1 + 4k^2+2k + 4k^2 )$
$S = 1 + \sum_{k=1}^7 ( 12k^2 + 6k + 1 )$
$S = 1 + 12\frac{7(8)(15)}{6} + 6\frac{7(8)}{2} + 7$
$S = 1 + 2(840) + 3(56) + 7 = 1 + 1680 + 168 + 7 = 1856$.