Evaluate the Summation sum from x=1 to 5 of 2x+3
The problem provided is asking for the calculation of a finite sum. Specifically, it requests to evaluate the series that results from plugging in numbers 1 through 5 into the formula 2x+3 for each term and then summing all those terms together. Essentially, you're calculating the sum of a sequence of arithmetic operations applied to the first 5 positive integers.
$\sum_{x = 1}^{5} 2 x + 3$
Solution:
Step 1: Write out the full series by substituting $x$ with each integer from 1 to 5.
$2 \cdot 1 + 3, 2 \cdot 2 + 3, 2 \cdot 3 + 3, 2 \cdot 4 + 3, 2 \cdot 5 + 3$
Step 2: Calculate the sum of the terms in the series.
The total is $45$.
"Step 1: Detail the sequence by inserting the values from 1 to 5 into the expression $2x + 3$.
$2 \times 1 + 3, 2 \times 2 + 3, 2 \times 3 + 3, 2 \times 4 + 3, 2 \times 5 + 3$
Step 2: Add up the values of the sequence to find the sum.
The sum is $45$."
The problem asks for the evaluation of a summation, which is a mathematical notation used to represent the addition of a sequence of numbers. The summation in question is $\sum_{x=1}^{5} (2x + 3)$, which means we need to calculate the sum of the expression $2x + 3$ as $x$ takes on each integer value from 1 to 5.
To solve this, we follow these steps:
Expansion: We expand the summation by computing the expression $2x + 3$ for each value of $x$ from 1 to 5. This step turns the summation into a series of individual terms.
Simplification: We then add these terms together to find the total sum. This is a straightforward arithmetic operation.
In the context of algebra and arithmetic, summation is a fundamental concept that allows for the concise representation of adding up a list of numbers. The sigma notation ($\sum$) is used to denote the summation, with the variable of summation (in this case, $x$) and the lower and upper bounds of summation (1 and 5, respectively) specified. The process of evaluating a summation involves finding the value of each term within the bounds and then summing these values.