Question

Find each sum. $$ \sum_{n=1}^{80}(4 n-9) $$

Short answer

12240

Step-by-step solution

- Understand the Summation Notation

The given expression is a sum of the terms from 1 to 80, of the form \(4n - 9\). This can be written as follows: \[ \sum_{n=1}^{80}(4n-9) \]

- Separate the Summation

Use the linearity of summation to break this into two separate sums: \[ \sum_{n=1}^{80}(4n-9) = \sum_{n=1}^{80} 4n - \sum_{n=1}^{80} 9 \]

- Factor Out Constants

Factor out the constants in each summation: \[ \sum_{n=1}^{80} 4n - \sum_{n=1}^{80} 9 = 4 \sum_{n=1}^{80} n - 9 \sum_{n=1}^{80} 1 \]

- Use Summation Formulas

Apply the formula for the sum of the first \(k\) integers: \[ \sum_{n=1}^k n = \frac{k(k+1)}{2} \] and the formula for the sum of 1 added \(k\) times: \[ \sum_{n=1}^k 1 = k \] Here, \(k = 80\).

- Substitute and Simplify

Substitute \(k=80\) into the formulas: \[ 4 \sum_{n=1}^{80} n - 9 \sum_{n=1}^{80} 1 = 4 \cdot \frac{80 \cdot 81}{2} - 9 \cdot 80 \] Simplify the expression: \[ 4 \cdot 3240 - 720 = 12960 - 720 = 12240 \]

Key concepts

sum of integers

The sum of integers up to a specific number can be calculated using a simple formula. This formula is particularly useful when dealing with large sums. For instance, the sum of the first 80 integers is found using the formula: \[ \text{Sum} = \frac{k(k+1)}{2} \] Here, \(k\) is the largest integer in the sum. By substituting \(k=80\), we get: \[ \frac{80 \times 81}{2} = 3240 \] Thus, the sum of the first 80 integers is 3240. This formula greatly simplifies the process of summing a consecutive series of integers.

linearity of summation

Linearity of summation is a powerful property. It states that the sum of a combined function can be separated into individual sums. For example, the sum \( \textstyle \sum_{n=1}^{80} (4n - 9) \) can be split into two sums: \[ \textstyle \sum_{n=1}^{80} 4n - \sum_{n=1}^{80} 9 \] This property simplifies the summation process. Additionally, it allows us to handle each part of the expression separately. This separation is crucial for making complex sums more manageable and for applying known formulas to each part individually.

factoring constants

When dealing with summation, factoring constants helps simplify calculations. For example, in the sum \( \textstyle \sum_{n=1}^{80} 4n \), the constant 4 can be factored out: \[ 4 \sum_{n=1}^{80} n \] Similarly, for the sum \( \textstyle \sum_{n=1}^{80} 9 \), we can factor out the 9: \[ 9 \sum_{n=1}^{80} 1 \] This step is key because it reduces the complexity of the expressions. It allows us to apply summation formulas to simpler components, making the overall problem easier to solve.