Chapter 13: Problem 54
Find each sum. $$ \sum_{n=1}^{80}\left(\frac{1}{3} n+\frac{1}{2}\right) $$
Short Answer
Expert verified
The sum is 1120.
Step by step solution
01
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.
02
Split the Summation
Rewrite the summation by splitting it into two separate sums: \(\sum_{n=1}^{80} \frac{1}{3} n + \sum_{n=1}^{80} \frac{1}{2}\).
03
Factor Out Constants
Factor out the constants from the sums: \(\frac{1}{3} \sum_{n=1}^{80} n + \frac{1}{2} \sum_{n=1}^{80} 1\).
04
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.
05
Combine the Results
Substitute these results back into the expression: \(\frac{1}{3} \times 3240 + \frac{1}{2} \times 80\).
06
Simplify
Simplify the expression: \(\frac{1}{3} \times 3240 = 1080\) and \(\frac{1}{2} \times 80 = 40\), so the final sum is \(1080 + 40 = 1120\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
\[ \sum_{i=1}^{k} i = \frac{k(k+1)}{2} \]
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:
\[ \sum_{i=1}^{k} c = k \times c \]
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.
\[ S_n = \frac{n}{2} (a + l) \]
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.