Problem 74 - A Graphical Approach to Precalculus with Limits

Question

Find the sum for each series. $$\sum_{i=1}^{52}\left(i^{2}+27 i+180\right)$$

Step-by-Step Solution

Answer

The sum is 93992.

Step 1: Identify the Series Expression

The series given is \(\sum_{i=1}^{52}(i^2 + 27i + 180)\). This means we need to add up the expression \(i^2 + 27i + 180\) for each integer \(i\) from 1 to 52.

Step 2: Separate the Series into Components

We can separate the terms in the sum into individual series: \(\sum_{i=1}^{52}i^2\), \(\sum_{i=1}^{52}27i\), and \(\sum_{i=1}^{52}180\). This allows us to handle each component separately.

Step 3: Compute Sum of \(i^2\)

The sum of squares formula is \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\). Plugging in \(n = 52\), we get: \(\frac{52 \times 53 \times 105}{6}\). Compute this to find the result.

Step 4: Compute Sum of \(27i\)

Use the arithmetic series formula \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\). Here, multiply the result by 27: \(27 \times \frac{52 \times 53}{2}\). Compute to find the result.

Step 5: Compute Sum of Constant Term

For a constant \(c\) added over \(n\) terms, the sum is \(c \times n\). Here, \(180 \times 52\). Compute to find the result.

Step 6: Combine the Results

Add together all the results from steps 3, 4, and 5 to obtain the total sum of the original series.

Step 7: Final Calculation

Now carry out the calculations:1. \(\sum_{i=1}^{52}i^2 = \frac{52 \times 53 \times 105}{6} = 47426\)2. \(\sum_{i=1}^{52}27i = 27 \times \frac{52 \times 53}{2} = 37206\)3. \(\sum_{i=1}^{52}180 = 180 \times 52 = 9360\)Add them: \(47426 + 37206 + 9360 = 93992\).

Key Concepts

Sum of Squares

The sum of squares is a fundamental concept in mathematics where we add together the squares of a sequence of numbers. In our exercise, the sequence ranges from 1 to 52. The formula for calculating the sum of squares is given by:

• \[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]

To understand this, let's break it down:- **\(n\)** represents the last number in the sequence, which is 52 in our case.- The three components of the formula reflect the relationships between successive integers. This formula calculates the sum of squares efficiently by simplifying the process of adding each squared number individually.Using the formula, where \(n = 52\):\[\sum_{i=1}^{52} i^2 = \frac{52 \times 53 \times 105}{6}\]Compute that to find the result of 47426.It's a neat formula that saves a lot of time, especially over large series. Sum of squares is essential not only in theoretical mathematics but also in statistics, where it helps describe variance and standard deviation.

Arithmetic Series

An arithmetic series is the sum of terms in an arithmetic sequence, where each term after the first is generated by adding a fixed, non-zero number (known as the 'common difference').

In this exercise, the middle term of the given series consists of an arithmetic sequence with a common difference of one:

• \[ \sum_{i=1}^{52} 27i \]

The sum of the first \(n\) integers can be computed using the formula:

• \[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]

Then you multiply by 27, as each term in the series is multiplied by this constant. So, the computation becomes:\[27 \times \frac{52 \times 53}{2}\]Carrying out the computation gives us the result of 37206. This step efficiently sums up all the terms produced by multiplying each integer in the range by 27. Understanding the arithmetic series formula can greatly enhance efficiency and comprehension when dealing with sequences and their sums.

Constant Term Sum

The constant term sum is often the simplest component to compute in a series. When you have a constant term repeated across a specified number of terms, the computation is straightforward.

In our exercise, let's examine the term:

• \[ \sum_{i=1}^{52} 180 \]

This states that the number 180 is summed across the series from 1 to 52. The formula is:

• \[ \text{Sum} = c \times n \]

Where \(c\) is the constant and \(n\) is the number of terms. Plugging in the numbers:\[180 \times 52 = 9360\]This step is vital in series calculations since it allows for the easy computation of any constant values added repeatedly through the series. Summation of constant terms simplifies the process of computing the overall total of the parts of the series. It's widely applicable not just in mathematics but also in real-world scenarios where consistent values are aggregated over periods or items.