Evaluate the Summation sum from i=4 to 12 of -4-6i
The question is asking for a calculation of a mathematical series. Specifically, it is requesting to evaluate the sum of terms from the fourth term to the twelfth term, where each term is defined by the expression -4-6i. The variable "i" represents the index of each term in the series. Therefore, you are meant to calculate the sum by plugging in the values of i from 4 through 12 into the expression, calculating each term individually, and then adding all of those terms together to find the total sum.
$\sum_{i = 4}^{12} - 4 - 6 i$
Answer
Solution:
Step 1:
Transform the original summation to start from $i=1$ by subtracting the unwanted terms.
$$\sum_{i = 4}^{12} (-4 - 6i) = \sum_{i = 1}^{12} (-4 - 6i) - \sum_{i = 1}^{3} (-4 - 6i)$$
Step 2:
Calculate $\sum_{i = 1}^{12} (-4 - 6i)$.
Step 2.1:
Decompose the summation into two separate summations.
$$\sum_{i = 1}^{12} (-4 - 6i) = \sum_{i = 1}^{12} (-4) + \sum_{i = 1}^{12} (-6i)$$
Step 2.2:
Compute $\sum_{i = 1}^{12} (-4)$.
Step 2.2.1:
Use the formula for the sum of a constant term.
$$\sum_{i = 1}^{n} c = cn$$
Step 2.2.2:
Insert the specific values into the formula.
$$(-4)(12)$$
Step 2.2.3:
Perform the multiplication.
$$-48$$
Step 2.3:
Calculate $-6 \sum_{i = 1}^{12} i$.
Step 2.3.1:
Apply the formula for the sum of the first $n$ natural numbers.
$$\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$$
Step 2.3.2:
Substitute the numbers into the formula and include the multiplier.
$$-6 \left( \frac{12(12 + 1)}{2} \right)$$
Step 2.3.3:
Simplify the expression.
Step 2.3.3.1:
Add $12 + 1$.
$$-6 \left( \frac{12 \cdot 13}{2} \right)$$
Step 2.3.3.2:
Eliminate the common factor of $2$.
Step 2.3.3.2.1:
Extract $2$ from $-6$.
$$2(-3) \frac{156}{2}$$
Step 2.3.3.2.2:
Cancel out the $2$s.
$$\cancel{2} \cdot (-3) \frac{156}{\cancel{2}}$$
Step 2.3.3.2.3:
Rewrite the simplified expression.
$$-3 \cdot 156$$
Step 2.3.3.3:
Multiply $-3$ by $156$.
$$-468$$
Step 2.4:
Combine the results of the two summations.
$$-48 - 468$$
Step 2.5:
Subtract $468$ from $-48$.
$$-516$$
Step 3:
Determine $\sum_{i = 1}^{3} (-4 - 6i)$.
Step 3.1:
Write out the terms for each value of $i$.
$$-4 - 6 \cdot 1 - 4 - 6 \cdot 2 - 4 - 6 \cdot 3$$
Step 3.2:
Simplify the series.
Step 3.2.1:
Multiply $-6$ by $1$.
$$-4 - 6 - 4 - 6 \cdot 2 - 4 - 6 \cdot 3$$
Step 3.2.2:
Combine $-6$ and $-4$.
$$-10 - 4 - 6 \cdot 2 - 4 - 6 \cdot 3$$
Step 3.2.3:
Multiply $-6$ by $2$.
$$-10 - 4 - 12 - 4 - 6 \cdot 3$$
Step 3.2.4:
Combine $-12$ and $-4$.
$$-10 - 16 - 4 - 6 \cdot 3$$
Step 3.2.5:
Combine $-16$ and $-10$.
$$-26 - 4 - 6 \cdot 3$$
Step 3.2.6:
Multiply $-6$ by $3$.
$$-26 - 4 - 18$$
Step 3.2.7:
Combine $-18$ and $-4$.
$$-26 - 22$$
Step 3.2.8:
Combine $-22$ and $-26$.
$$-48$$
Step 4:
Substitute the calculated values back into the original equation.
$$-516 + 48$$
Step 5:
Add $-516$ and $48$.
$$-468$$
Knowledge Notes:
The problem involves evaluating a summation of a linear polynomial in the form of $-4-6i$ from $i=4$ to $i=12$. The process involves several mathematical concepts:
Summation Notation: The use of the sigma symbol ($\Sigma$) to denote the sum of a sequence of terms defined by an expression inside the summation.
Arithmetic Series: A sequence of numbers in which each term after the first is obtained by adding a constant to the previous term. The sum of an arithmetic series can be calculated using specific formulas.
Sum of a Constant: The sum of a constant $c$ over $n$ terms is simply $cn$.
Sum of the First $n$ Natural Numbers: The sum of the first $n$ natural numbers is given by the formula $\frac{n(n+1)}{2}$.
Algebraic Manipulation: The process involves algebraic manipulation, such as distributing a constant across a summation, simplifying expressions, and combining like terms.
Properties of Summation: The linearity of summation allows for the separation and combination of summations, which is used to break down the original problem into more manageable parts.
By applying these concepts, the problem is systematically broken down into smaller parts, each of which is solved using the appropriate formulas and algebraic techniques. The final result is obtained by combining the individual results of the summations.
