Chapter 13: Problem 77
Find the sum of each sequence. \(\sum_{k=10}^{60}(2 k)\)
Short Answer
Expert verified
The sum is 3570.
Step by step solution
01
Identify the Sequence
The given sequence is \(\for_all_{k=10}^{60} 2k\). This represents the sum of multiples of 2 from k=10 to k=60.
02
Set Up the Summation
Summations can be simplified by factoring out constants. Rewrite the sequence as \(\for_all_{k=10}^{60} 2k = 2 \for_all_{k=10}^{60} k\). This allows focusing on summing the integers from 10 to 60.
03
Sum of Integers from k=10 to k=60
To sum the integers from k=10 to k=60, use the formula for the sum of an arithmetic series: \[S = \frac{n}{2} (a + l)\] where n is the number of terms, a is the first term, and l is the last term. Here, \(a = 10\), \(l = 60\), and \(n = (60 - 10 + 1) = 51\).
04
Calculate the Sum of Integers
Apply the arithmetic series sum formula: \[S = \frac{51}{2} (10 + 60) = \frac{51}{2} \times 70 = 51 \times 35 = 1785\]
05
Include the Factor 2
Remember to multiply the result by 2, as factored earlier: \[2 \times 1785 = 3570\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
arithmetic series
An arithmetic series is the sum of the terms in an arithmetic sequence. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This difference is known as the common difference. For example, in the sequence 2, 4, 6, 8, the common difference is 2.
When dealing with arithmetic series, the goal is to find the sum of all terms in the sequence.
This is often done using a special formula for summation which makes the calculation easier and faster.
When dealing with arithmetic series, the goal is to find the sum of all terms in the sequence.
This is often done using a special formula for summation which makes the calculation easier and faster.
sequence summation
Sequence summation involves adding a series of numbers together. In our example, we needed to find the sum of the sequence \(\for_all_{k=10}^{60} 2k\).
To do this, we first recognized that 2 is a constant factor and can be 'factored out' to simplify our work. This left us with a simpler problem, focusing on summing the integers from 10 to 60. Summation of sequences is a fundamental concept in mathematics, allowing for the concise combination of many terms into a single sum.
To do this, we first recognized that 2 is a constant factor and can be 'factored out' to simplify our work. This left us with a simpler problem, focusing on summing the integers from 10 to 60. Summation of sequences is a fundamental concept in mathematics, allowing for the concise combination of many terms into a single sum.
factorizing constants
When you factorize constants in a sequence summation, you simplify the sum by temporarily removing the constant. In the example \(2 \for_all_{k=10}^{60} k\), the constant 2 is taken outside the summation symbol. This leaves us with the easier task of summing just the integers from 10 to 60.
After completing the summation of the integers, the final result is then multiplied by the factored-out constant to get the total sum. This efficiency saves time and reduces the possibility of calculation errors.
After completing the summation of the integers, the final result is then multiplied by the factored-out constant to get the total sum. This efficiency saves time and reduces the possibility of calculation errors.
sum formula
The sum formula for an arithmetic series is key to quickly finding the sum of a sequence without adding each term individually. The formula is: \[S = \frac{n}{2} (a + l)\]
Here, \(S\) is the sum, \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term. In our case, \(a = 10\), \(l = 60\), and \(n = 51\).
Substituting these values into the formula, we get: \[S = \frac{51}{2} \times (10 + 60) = 35 \times 51 = 1785\]
This sum is then multiplied by the previously factored 2, yielding the final result of 3570.
Here, \(S\) is the sum, \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term. In our case, \(a = 10\), \(l = 60\), and \(n = 51\).
Substituting these values into the formula, we get: \[S = \frac{51}{2} \times (10 + 60) = 35 \times 51 = 1785\]
This sum is then multiplied by the previously factored 2, yielding the final result of 3570.