An animal hospital advertises a veterinarian's starting salary as $50,000 for the first year, with increases of $1,500 each year for the next nineteen years.
The total salary is represented by the series shown.
[tex]\sum_{k=1}^{20}[50,000+(k-1) 1,500][/tex]
Which is equivalent to the series? Check all that apply.
[tex]\sum_{k=1}^{20} 50,000+\sum_{k=1}^{20}(k-1) 1,500[/tex]
[tex]\sum_{k=1}^{20} 50,000+1,500 \sum_{k=1}^{20}(k-1)[/tex]
[tex]50,000 \sum_{k=1}^{20} k+1,500 \sum_{k=1}^{20}(k-1)[/tex]
[tex]\sum_{k=1}^{20} 50,000+1,500 \sum_{k=1}^{20} k-\sum_{k=1}^{20} 1,500[/tex]
Answer (2)
Apply the summation property to split the original series: ∑ k = 1 20 [ 50 , 000 + ( k − 1 ) 1 , 500 ] = ∑ k = 1 20 50 , 000 + ∑ k = 1 20 ( k − 1 ) 1 , 500 .
Apply the constant multiple property: ∑ k = 1 20 50 , 000 + ∑ k = 1 20 ( k − 1 ) 1 , 500 = ∑ k = 1 20 50 , 000 + 1 , 500 ∑ k = 1 20 ( k − 1 ) .
Rewrite the fourth option and verify its equivalence: ∑ k = 1 20 50 , 000 + 1 , 500 ∑ k = 1 20 k − ∑ k = 1 20 1 , 500 .
The equivalent series are: k = 1 ∑ 20 50 , 000 + k = 1 ∑ 20 ( k − 1 ) 1 , 500 , k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 ( k − 1 ) , k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 k − k = 1 ∑ 20 1 , 500
Explanation
Understanding the Problem We are given the series ∑ k = 1 20 [ 50 , 000 + ( k − 1 ) 1 , 500 ] representing the total salary of a veterinarian over 20 years. The starting salary is $50,000, and it increases by $1,500 each year for the next 19 years. Our goal is to identify which of the given series are equivalent to the original series.
Applying Summation Property We can use the summation property ∑ k = 1 n ( a k + b k ) = ∑ k = 1 n a k + ∑ k = 1 n b k to split the original series into two separate summations: k = 1 ∑ 20 [ 50 , 000 + ( k − 1 ) 1 , 500 ] = k = 1 ∑ 20 50 , 000 + k = 1 ∑ 20 ( k − 1 ) 1 , 500 This matches the first option.
Applying Constant Multiple Property Next, we can use the constant multiple property ∑ k = 1 n c ⋅ a k = c ∑ k = 1 n a k to factor out the 1,500 from the second summation: k = 1 ∑ 20 50 , 000 + k = 1 ∑ 20 ( k − 1 ) 1 , 500 = k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 ( k − 1 ) This matches the second option.
Analyzing the Third Option The third option is 50 , 000 ∑ k = 1 20 k + 1 , 500 ∑ k = 1 20 ( k − 1 ) . This is not equivalent because the original series has a constant term of 50,000 summed 20 times, not a term that is multiplied by k.
Analyzing the Fourth Option Now, let's analyze the fourth option: ∑ k = 1 20 50 , 000 + 1 , 500 ∑ k = 1 20 k − ∑ k = 1 20 1 , 500 . We can rewrite ∑ k = 1 20 ( k − 1 ) as ∑ k = 1 20 k − ∑ k = 1 20 1 . Then, 1 , 500 k = 1 ∑ 20 ( k − 1 ) = 1 , 500 ( k = 1 ∑ 20 k − k = 1 ∑ 20 1 ) = 1 , 500 k = 1 ∑ 20 k − 1 , 500 k = 1 ∑ 20 1 So, ∑ k = 1 20 50 , 000 + 1 , 500 ∑ k = 1 20 k − ∑ k = 1 20 1 , 500 is equivalent to ∑ k = 1 20 50 , 000 + ∑ k = 1 20 ( k − 1 ) 1 , 500 . This matches the fourth option.
Final Answer Therefore, the equivalent series are: k = 1 ∑ 20 50 , 000 + k = 1 ∑ 20 ( k − 1 ) 1 , 500 k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 ( k − 1 ) k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 k − k = 1 ∑ 20 1 , 500
Examples
Understanding series and summations is very useful in calculating total costs or revenues over a period. For example, if a company's revenue increases by a fixed amount each year, you can use a series to calculate the total revenue over several years. Similarly, if you are saving money and depositing a fixed amount into an account each month, you can use a series to calculate the total amount saved over time. These concepts are fundamental in financial planning and business analysis.
The series k = 1 ∑ 20 [ 50 , 000 + ( k − 1 ) 1 , 500 ] can be broken down into several equivalent series. The equivalent forms are: k = 1 ∑ 20 50 , 000 + k = 1 ∑ 20 ( k − 1 ) 1 , 500 , k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 ( k − 1 ) , and k = 1 ∑ 20 50 , 000 + 1 , 500 k = 1 ∑ 20 k − k = 1 ∑ 20 1 , 500 .
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