Anderson earns $20 per month for completing chores around the house. Each month, the amount Anderson earns increases by $0.50. The total amount of money Anderson earns from month 3 to month 18 can be represented by this expression: [tex]$\sum_{n=3}^{18}[20+(n-1) 0.5]$[/tex]. Which is another way of expressing this amount?
[tex]$\sum_{n=1}^{18} 20+0.5 \sum_{n=1}^{18} n-0.5 \sum_{n=1}^{18} 1-\left(\sum_{n=1}^2 20+0.5 \sum_{n=1}^2 n-0.5 \sum_{n=1}^2 1\right)$[/tex]
[tex]$\sum_{n=1}^{18} 20+0.5 \sum_{n=1}^{18} n-0.5 \sum_{n=1}^{18} 1-\left(\sum_{n=1}^3 20+0.5 \sum_{n=1}^3 n-0.5 \sum_{n=1}^3 1\right)$[/tex]
[tex]$\sum_{n=1}^{18} 20+0.5 \sum_{n=1}^{18} n-\left(\sum_{n=1}^2 20+0.5 \sum_{n=1}^2 n\right)$[/tex]
[tex]$\sum_{\sum 20}^{18}+0.5 \sum_{n-}^{18}\left(\sum^3 20+0.5 \sum_n^3\right)$[/tex]
Answer (2)
Expand the original summation.
Rewrite the summation to start from n=1.
Substitute and simplify the expression.
The equivalent expression is n = 1 ∑ 18 20 + 0.5 n = 1 ∑ 18 n − 0.5 n = 1 ∑ 18 1 − ( n = 1 ∑ 2 20 + 0.5 n = 1 ∑ 2 n − 0.5 n = 1 ∑ 2 1 ) .
Explanation
Understanding the Problem We are given the expression ∑ n = 3 18 [ 20 + ( n − 1 ) 0.5 ] which represents the total amount of money Anderson earns from month 3 to month 18. We need to find an equivalent expression from the given options.
Expanding the Summation Let's expand the given summation: n = 3 ∑ 18 [ 20 + ( n − 1 ) 0.5 ] = n = 3 ∑ 18 20 + n = 3 ∑ 18 ( n − 1 ) 0.5 = n = 3 ∑ 18 20 + 0.5 n = 3 ∑ 18 ( n − 1 ) .
Rewriting the Summation Now, let's rewrite the summation to start from n = 1 . We can express the sum from n = 3 to 18 as the sum from n = 1 to 18 minus the sum from n = 1 to 2 . So, we have n = 3 ∑ 18 20 = n = 1 ∑ 18 20 − n = 1 ∑ 2 20 and n = 3 ∑ 18 ( n − 1 ) = n = 1 ∑ 18 ( n − 1 ) − n = 1 ∑ 2 ( n − 1 ) .
Expanding the Second Term Expanding the second term, we get n = 1 ∑ 18 ( n − 1 ) = n = 1 ∑ 18 n − n = 1 ∑ 18 1 and n = 1 ∑ 2 ( n − 1 ) = n = 1 ∑ 2 n − n = 1 ∑ 2 1 .
Substituting Back Substituting back into the original expression, we have n = 1 ∑ 18 20 − n = 1 ∑ 2 20 + 0.5 ( n = 1 ∑ 18 n − n = 1 ∑ 18 1 ) − 0.5 ( n = 1 ∑ 2 n − n = 1 ∑ 2 1 ) = n = 1 ∑ 18 20 + 0.5 n = 1 ∑ 18 n − 0.5 n = 1 ∑ 18 1 − ( n = 1 ∑ 2 20 + 0.5 n = 1 ∑ 2 n − 0.5 n = 1 ∑ 2 1 ) .
Finding the Equivalent Expression Comparing the derived expression with the given options, we find that the equivalent expression is n = 1 ∑ 18 20 + 0.5 n = 1 ∑ 18 n − 0.5 n = 1 ∑ 18 1 − ( n = 1 ∑ 2 20 + 0.5 n = 1 ∑ 2 n − 0.5 n = 1 ∑ 2 1 ) .
Examples
Understanding series and summations is crucial in many fields, such as physics and engineering. For example, when calculating the total energy of a system with discrete energy levels, we often use summations to add up the energy of each level. Similarly, in finance, we use summations to calculate the total return on an investment over a period of time. These concepts help us model and analyze complex systems by breaking them down into smaller, manageable parts.
The equivalent expression for the total amount Anderson earns from month 3 to month 18 is n = 1 ∑ 18 20 + 0.5 n = 1 ∑ 18 n − 0.5 n = 1 ∑ 18 1 − ( n = 1 ∑ 2 20 + 0.5 n = 1 ∑ 2 n − 0.5 n = 1 ∑ 2 1 ) . This captures all earnings accurately by adjusting the range of summation and breaking down the terms step by step.
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