Evaluate the Summation sum from k=5 to 20 of k^3

Solution:

Step 1:

Adjust the summation to start from $k=1$ instead of $k=5$.

$$\sum _{k=5}^{20}{k}^3=\sum _{k=1}^{20}{k}^3-\sum _{k=1}^4{k}^3$$

Step 2:

Calculate the sum from $k=1$ to $k=20$ using the cubic sum formula.

Step 2.1:

Utilize the formula for the sum of cubes up to $n$:

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

Step 2.2:

Insert $n=20$ into the formula.

$${\left(\frac{20(20+1)}{2}\right)}^2$$

Step 2.3:

Perform the simplification.

Step 2.3.1:

First, simplify the numerator.

Step 2.3.1.1:

Combine $20$ and $1$.

$${\left(\frac{20\cdot 21}{2}\right)}^2$$

Step 2.3.1.2:

Square $20$.

$${\left(\frac{400\cdot 21}{2}\right)}^2$$

Step 2.3.1.3:

Square $21$.

$${\left(\frac{400\cdot 441}{2}\right)}^2$$

Step 2.3.2:

Now, simplify the entire expression.

Step 2.3.2.1:

Multiply $400$ by $441$.

$${\left(\frac{176400}{2}\right)}^2$$

Step 2.3.2.2:

Divide $176400$ by $2$.

$${88200}^2$$

Step 3:

Determine the sum from $k=1$ to $k=4$.

Step 3.1:

Write out the series for each $k$ value.

$$1^3+2^3+3^3+4^3$$

Step 3.2:

Simplify the series.

Step 3.2.1:

Cube $1$.

$$1+2^3+3^3+4^3$$

Step 3.2.2:

Cube $2$.

$$1+8+3^3+4^3$$

Step 3.2.3:

Combine $1$ and $8$.

$$9+3^3+4^3$$

Step 3.2.4:

Cube $3$.

$$9+27+4^3$$

Step 3.2.5:

Add $9$ and $27$.

$$36+4^3$$

Step 3.2.6:

Cube $4$.

$$36+64$$

Step 3.2.7:

Combine $36$ and $64$.

$$100$$

Step 4:

Substitute the calculated sums back into the adjusted summation.

$${88200}^2-100$$

Step 5:

Subtract $100$ from ${88200}^2$.

$${88200}^2-100$$

Knowledge Notes:

The problem involves evaluating a finite summation of cubic terms. The key knowledge points for solving this problem are:

1. Summation Notation: The sigma notation $\sum$ is used to denote the sum of a sequence of numbers. The expression $\sum _{k=a}^{b}f(k)$ represents the sum of the function $f(k)$ evaluated at all integers from $a$ to $b$ inclusive.

2. Sum of Cubes Formula: The sum of the first $n$ cubes, $\sum _{k=1}^n{k}^3$, can be calculated using the formula ${\left(\frac{n(n+1)}{2}\right)}^2$. This formula is derived from the properties of arithmetic progressions and algebraic identities.

3. Arithmetic Operations: Basic arithmetic operations, such as addition, multiplication, exponentiation, and simplification, are used throughout the problem to evaluate the summation.

4. Adjusting the Limits of Summation: To evaluate a summation starting from a number other than $1$, we can adjust the limits by subtracting the sum from $1$ up to the number just before the new starting point.

5. Direct Computation for Small Series: For small values of $n$, it may be more efficient to compute the sum directly rather than using the sum of cubes formula.

By understanding and applying these concepts, the problem can be solved systematically, step by step, to find the sum of cubic numbers from $k=5$ to $k=20$.