Evaluate the Summation sum from i=1 to 6 of 3i^3

Solution:

Step 1:

Extract the constant $3$ from the summation: $3 \sum_{i = 1}^{6} i^{3}$

Step 2:

Apply the cubic summation formula: $\sum_{i = 1}^{n} i^{3} = \left(\frac{n(n + 1)}{2}\right)^{2}$

Step 3:

Insert $n = 6$ into the formula and include the extracted constant: $3 \times \left(\frac{6(6 + 1)}{2}\right)^{2}$

Step 4:

Begin simplification:

Step 4.1:

Start with the numerator:

Step 4.1.1:

Combine $6$ and $1$: $3 \times \frac{6^{2} \times 7^{2}}{4}$

Step 4.1.2:

Square $6$: $3 \times \frac{36 \times 7^{2}}{4}$

Step 4.1.3:

Square $7$: $3 \times \frac{36 \times 49}{4}$

Step 4.2:

Continue simplifying the expression:

Step 4.2.1:

Calculate $36 \times 49$: $3 \times \left(\frac{1764}{4}\right)$

Step 4.2.2:

Divide $1764$ by $4$: $3 \times 441$

Step 4.2.3:

Multiply $3$ by $441$: $1323$

Knowledge Notes:

The problem involves evaluating a summation of a cubic polynomial. The key knowledge points include:

1. Summation Notation: The summation notation $\sum$ is used to denote the sum of a sequence of terms. The expression $\sum_{i=1}^{n} f(i)$ means to sum the function $f(i)$ as $i$ goes from $1$ to $n$.

2. Factoring Constants: In a summation, constants can be factored out and multiplied after the summation is evaluated. This simplifies the process by reducing the complexity of the terms being summed.

3. Cubic Summation Formula: The formula for the sum of the first $n$ cubes is given by $\sum_{i=1}^{n} i^{3} = \left(\frac{n(n + 1)}{2}\right)^{2}$. This formula allows for quick calculation without having to sum each term individually.

4. Arithmetic Operations: Basic arithmetic operations such as addition, multiplication, and exponentiation are used to simplify the expression.

5. Simplification: The process of simplification involves performing arithmetic operations in the correct order (according to the order of operations) to reduce the expression to its simplest form.