Problem 54 - Algebra and Trigonometry

Question

Find each sum. $$ \sum_{n=1}^{80}\left(\frac{1}{3} n+\frac{1}{2}\right) $$

Step-by-Step Solution

Answer

The sum is 1120.

Step 1: Understand the Summation Notation

The expression \(\sum_{n=1}^{80}\left(\frac{1}{3} n+\frac{1}{2}\right)\) represents the sum of the terms \(\frac{1}{3} n+\frac{1}{2}\) as n ranges from 1 to 80.

Step 2: Split the Summation

Rewrite the summation by splitting it into two separate sums: \(\sum_{n=1}^{80} \frac{1}{3} n + \sum_{n=1}^{80} \frac{1}{2}\).

Step 3: Factor Out Constants

Factor out the constants from the sums: \(\frac{1}{3} \sum_{n=1}^{80} n + \frac{1}{2} \sum_{n=1}^{80} 1\).

Step 4: Evaluate the Summations

The sum \(\sum_{n=1}^{80} n\) is given by the formula for the sum of the first n natural numbers: \(\sum_{n=1}^{k} n = \frac{k(k+1)}{2}\). So, \(\sum_{n=1}^{80} n = \frac{80(80+1)}{2} = 3240\). The sum of constants \(\sum_{n=1}^{80} 1\) is just 80.

Step 5: Combine the Results

Substitute these results back into the expression: \(\frac{1}{3} \times 3240 + \frac{1}{2} \times 80\).

Step 6: Simplify

Simplify the expression: \(\frac{1}{3} \times 3240 = 1080\) and \(\frac{1}{2} \times 80 = 40\), so the final sum is \(1080 + 40 = 1120\).

Key Concepts

sum of natural numbers

Summing natural numbers is a common process in mathematics, often denoted using summation notation. Understanding this is crucial in solving our exercise. The sum of the first n natural numbers is given by a well-known formula:

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

This formula simplifies the addition of a series of consecutive numbers. For example, if you want to sum the first 80 natural numbers ( from 1 to 80), substitute 80 in the formula:

\[ \sum_{i=1}^{80} i = \frac{80 \times (80+1)}{2} = 3240 \]

This formula helps us quickly evaluate sums without having to list and add each number individually.

constants in summation

Handling constants within summation notation can simplify complex series. A constant can be pulled out of the summation, allowing the terms to be summed independently. For instance, consider the summation of a term \sum_{i=1}^{k} c \: where \ c \ is a constant. As each term is the same, this simplifies to:

\[ \sum_{i=1}^{k} c = k \times c \]

In our exercise, we have two separate sums after splitting and factoring out the constants:

\[ \frac{1}{3} \sum_{n=1}^{80} n + \frac{1}{2} \sum_{n=1}^{80} 1 \]

Here, the sum of constants \sum_{n=1}^{80} 1 \ is just 80 because it's the sum of 1 added 80 times:

mathematical series

A mathematical series is the sum of the terms of a sequence. Series can be finite or infinite. In our case, we have a finite series since we are summing terms from 1 to 80. We use different techniques to evaluate series depending on their complexity. For instance, arithmetic series have a common difference between terms, and their sum can be found using:

\[ S_n = \frac{n}{2} (a + l) \]

where \ n \ is the number of terms, \ a \ is the first term, and \ l \ is the last term. In our exercise:

\[ S_n = \frac{1}{3} \sum_{n=1}^{80} n + \frac{1}{2} \sum_{n=1}^{80} 1 \]

We evaluated each part separately and then combined the results to find the total sum. These principles and formulas help significantly when dealing with larger series and summations.