Evaluate the Summation sum from k=1 to 7 of k^3

Solution:

Step 1:

Utilize the standard formula for the sum of cubes up to a certain number $n$: $\sum_{k = 1}^{n} k^{3} = \left(\frac{n(n + 1)}{2}\right)^2$

Step 2:

Insert the upper limit of the summation into the formula: $\left(\frac{7(7 + 1)}{2}\right)^2$

Step 3:

Proceed with the calculation.

Step 3.1:

First, address the numerator.

Step 3.1.1:

Combine $7$ and $1$: $\left(\frac{7 \cdot 8}{2}\right)^2$

Step 3.1.2:

Square $7$: $\left(\frac{49 \cdot 8}{2}\right)^2$

Step 3.1.3:

Square $8$: $\left(\frac{49 \cdot 64}{2}\right)^2$

Step 3.2:

Finalize the simplification.

Step 3.2.1:

Multiply $49$ by $64$: $\left(\frac{3136}{2}\right)^2$

Step 3.2.2:

Divide $3136$ by $2$: $784$

Knowledge Notes:

The problem involves calculating the sum of the cubes of the first seven natural numbers. To solve this, we use a known formula for the sum of cubes, which is derived from the sum of a series of natural numbers raised to the third power. The formula is:

$$ \sum_{k = 1}^{n} k^{3} = \left(\frac{n(n + 1)}{2}\right)^2 $$

This formula is a result of mathematical induction and can be proven using various methods, including algebraic manipulations and combinatorial arguments. The right side of the equation represents the square of the sum of the first $n$ natural numbers, which is itself a well-known formula:

$$ \frac{n(n + 1)}{2} $$

When we apply this formula to our problem, we substitute $n$ with $7$ and follow the order of operations: first, we handle the parentheses by adding and squaring, then we perform multiplication, and finally, we divide as per the formula's structure. The result gives us the sum of the cubes from $1$ to $7$.