Problem 53 - Contemporary Precalculus

Question

$$\text { Find the sum.}$$ $$\sum_{n=1}^{40} \frac{n+3}{6}$$

Step-by-Step Solution

Answer

Answer: The overall sum of the first 40 terms of the sequence is 157.

Step 1: Split the expression into two parts

We can split our given expression \(\frac{n+3}{6}\) into two parts: the first part being \(\frac{n}{6}\) and the second part being \(\frac{3}{6}\). Our sum then becomes $$\sum_{n=1}^{40} \frac{n}{6} + \sum_{n=1}^{40} \frac{3}{6}.$$

Step 2: Find the sum of the first part

We will first find the sum of the first part - $$\sum_{n=1}^{40} \frac{n}{6}$$. To do so, we can use the formula for the sum of the first n natural numbers, which is $$\frac{n(n+1)}{2}$$. In this case, we multiply each term by \(\frac{1}{6}\) to account for the \(\frac{n}{6}\) in the given expression. So, the sum of the first part will be $$\frac{1}{6} \cdot \frac{40(40+1)}{2}.$$

Step 3: Find the sum of the second part

Now we will find the sum of the second part - $$\sum_{n=1}^{40} \frac{3}{6}$$. Since the expression does not depend on n, we can simply multiply the \(\frac{3}{6}\) by the number of terms (40) to find the sum: $$\frac{3}{6} \cdot 40$$.

Step 4: Calculate the total sum

Now we have the sum of both parts: $$\frac{1}{6} \cdot \frac{40(40+1)}{2}$$ and $$\frac{3}{6} \cdot 40$$. We will now add these two sums together to find the total sum: $$\frac{1}{6} \cdot \frac{40(40+1)}{2} + \frac{3}{6} \cdot 40.$$ Calculating the numerical values we get $$\frac{1}{6} \cdot \frac{40 \cdot 41}{2} + \frac{1}{2} \cdot 40.$$ After some simplification: $$\frac{820}{6} + 20.$$ Finally, we find that the total sum is $$\boxed{157}.$$

Key Concepts

Arithmetic Series

An arithmetic series is a sequence of numbers in which each term after the first is formed by adding a constant difference to the previous term. This difference is known as the "common difference".
For example, in the sequence 3, 6, 9, 12, ..., the common difference is 3. Each term is obtained by adding 3 to the previous term.

• In an arithmetic series, you can sum up lots of numbers quickly using a formula.
• It's useful because instead of adding each number one by one, you use the properties of the series.

In our exercise, we transform an expression involving a series and break it down into simpler series components. By identifying the arithmetic nature, we efficiently compute the sum over a series of numbers. This process simplifies the task of finding the series' total sum.

Summation Formula

The summation formula for an arithmetic series greatly simplifies the addition of sequential numbers. It's a numerical shortcut.
The formula is given by:\[S_n = \frac{n}{2} (a + l)\]where:

• \(S_n\) is the sum of the series.
• \(n\) is the number of terms.
• \(a\) is the first term.
• \(l\) is the last term.

For a simple series from 1 to 40, this simplifies as \(a = 1\) and \(l = 40\). In most exercises, knowing this formula allows for quick calculations by simply plugging in the correct values. This is exactly what happens when you break an expression like the one given in our problem into partial sums and apply the basic summation formulas.

Natural Numbers

Natural numbers are the set of positive integers starting from 1, extending indefinitely (1, 2, 3, 4,...).
They are the foundation of arithmetic series. Calculations often revolve around finding sequences and sums over these numbers.

• They simplify mathematical operations, like the sum of numbers from 1 to \(n\).
• Used in the summation formula \(\frac{n(n+1)}{2}\) to find the sum of the first \(n\) natural numbers.

Whenever summation involves natural numbers—as in our exercise with numbers from 1 to 40—it highlights how effortlessly natural numbers help in structuring and breaking down more complex mathematical problems.

Mathematical Simplification

Mathematical simplification involves breaking down complex problems into more manageable parts. This is a crucial skill in algebra and arithmetic.
In our exercise solution, simplification is used to split the original series into two easier-to-sum parts.

• By expressing \(\frac{n+3}{6}\) as \(\frac{n}{6} + \frac{3}{6}\), the problem becomes much easier to handle for each summation part.
• This method allows for focused application of the summation formulas on each component.

Without simplification, solving complex series could become cumbersome. But, by simplifying a problem through strategic manipulation, we reach solutions faster and more efficiently. It reduces potential errors by tackling simpler parts rather than a tangled whole.