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 need to find the sum of the series $S = 1^3 - 2^3 + 3^3 - 4^3 + ... + 15^3$.
This is an alternating sum of cubes of the first 15 natural numbers.

Method 1: Expressing the alternating sum using the standard sum of cubes.
We know that $1^3 + 2^3 + 3^3 + ... + 15^3 = (1^3 + 3^3 + ... + 15^3) + (2^3 + 4^3 + ... + 14^3)$.
The alternating sum $S = (1^3 + 3^3 + ... + 15^3) - (2^3 + 4^3 + ... + 14^3)$.
Thus, $S = (1^3 + 2^3 + ... + 15^3) - 2(2^3 + 4^3 + ... + 14^3)$.

Step 1: Calculate the sum of the first 15 cubes.
Formula: $\sum_{r=1}^n r^3 = [\frac{n(n+1)}{2}]^2$.
For $n=15$: $\sum_{r=1}^{15} r^3 = [\frac{15 imes 16}{2}]^2 = (120)^2 = 14400$.

Step 2: Calculate $2(2^3 + 4^3 + ... + 14^3)$.
$2 \times 2^3 (1^3 + 2^3 + ... + 7^3) = 16 \times [\frac{7 imes 8}{2}]^2$
$= 16 \times (28)^2 = 16 \times 784 = 12544$.

Step 3: Find $S$.
$S = 14400 - 12544 = 1856$.