Evaluate the Summation sum from k=1 to 12 of 9
The question asks to compute the total sum of a series where the term "9" is repeatedly added for each value of "k" starting from k=1 until k=12. It's a straightforward arithmetic sequence where the value being summed does not change with each increment of "k". Essentially, you would be calculating the result of adding the number nine to itself twelve times.
$\sum_{k = 1}^{12} 9$
Identify the summation formula for a constant value, which is given by: $\sum_{k = 1}^{n} c = c \cdot n$
Insert the provided values into the established formula: $(9) \cdot (12)$
Calculate the product of $9$ and $12$: $108$
The problem at hand involves evaluating a simple summation where the term being summed is a constant. The relevant knowledge points for solving this problem include:
Summation Notation: Summation notation is a way to represent the addition of a sequence of numbers. The Greek letter sigma ($\Sigma$) is used to denote the summation.
Summation of a Constant: When summing a constant $c$ over $n$ terms, the result is simply the constant multiplied by the number of terms. This is because each term in the summation is the same, and adding $c$ to itself $n$ times is equivalent to multiplying $c$ by $n$.
Arithmetic Operations: Basic arithmetic operations are used in the process, specifically multiplication. Multiplying a number by an integer simply adds the number to itself that many times.
Substitution: This involves replacing the variables in an algebraic expression with their actual values. In this case, the constant $c$ is replaced with $9$, and $n$ is replaced with $12$.
Evaluation: The final step is to perform the arithmetic operation, which yields the answer to the summation problem.
The formula used in the solution is derived from the concept that the sum of a constant $c$ over $n$ terms is equal to $c$ times $n$. This is a fundamental principle in arithmetic and algebra that simplifies the process of summing constant values over a series.