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If \(\mathrm{x}\) is the sum of the first 50 positive even integers and \(\mathrm{y}\) is the sum of the first 50 positive odd integers, what is the value of \(\mathrm{x} - \mathrm{y}\) ?
Let's start by understanding exactly what we're looking for. The problem asks us to find two different sums:
What does "first 50 positive even integers" mean? These are: \(2, 4, 6, 8, 10, \ldots\), and we need exactly 50 of them. To find the 50th even integer, we use the pattern that the nth positive even integer is \(2n\). So the 50th even integer is \(2(50) = 100\).
Therefore: \(x = 2 + 4 + 6 + 8 + \ldots + 100\)
Similarly, "first 50 positive odd integers" means: \(1, 3, 5, 7, 9, \ldots\), and we need exactly 50 of them. The nth positive odd integer is \(2n-1\). So the 50th odd integer is \(2(50)-1 = 99\).
Therefore: \(y = 1 + 3 + 5 + 7 + \ldots + 99\)
We need to find \(x - y\).
Process Skill: TRANSLATE - Converting the problem language into clear mathematical understanding
Instead of calculating each sum separately (which would be very time-consuming), let's look for a smart pattern. Notice how even and odd integers relate to each other:
Do you see the pattern? Each even integer is exactly 1 more than its corresponding odd integer. This makes sense because consecutive integers differ by 1, and every even integer immediately follows an odd integer.
In general terms: the nth even integer minus the nth odd integer always equals 1.
Process Skill: INFER - Drawing the non-obvious conclusion that we can pair terms rather than calculate separate sums
Now we can use this pattern to find \(x - y\) without doing complex arithmetic:
\(x - y = (2 + 4 + 6 + 8 + \ldots + 100) - (1 + 3 + 5 + 7 + \ldots + 99)\)
We can rearrange this as:
\(x - y = (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7) + \ldots + (100 - 99)\)
Since each pair gives us a difference of 1:
\(x - y = 1 + 1 + 1 + 1 + \ldots + 1\)
How many 1's do we have? We have exactly 50 pairs (since we're pairing the 1st through 50th terms), so:
\(x - y = 50 \times 1 = 50\)
Our calculated answer is 50. Looking at the answer choices:
Our answer of 50 corresponds to choice C.
Let's do a quick sanity check with smaller numbers: If we took the first 3 even integers \((2, 4, 6)\) and first 3 odd integers \((1, 3, 5)\):
Using our method: 3 pairs, each with difference 1, so total difference = \(3 \times 1 = 3\). ✓
The answer is C. 50
The key insight was recognizing that each even integer is exactly 1 more than its corresponding odd integer, so when we have 50 such pairs, the total difference is \(50 \times 1 = 50\).
1. Misinterpreting "first 50 positive even integers"
Students often confuse this with "the first 50 multiples of 2" or think it means "all even integers up to 50." They might incorrectly identify the sequence as \(2, 4, 6, \ldots, 50\) instead of \(2, 4, 6, \ldots, 100\). This leads to completely wrong calculations from the start.
2. Attempting to calculate each sum separately using formulas
Many students immediately think to use arithmetic sequence formulas like \(S = n(\text{first term} + \text{last term})/2\) for each sum separately. While this works, it's much more time-consuming and error-prone than recognizing the elegant pairing pattern that makes this problem quick to solve.
3. Not recognizing the pairing opportunity
Students often miss that they can pair corresponding terms (1st even with 1st odd, 2nd even with 2nd odd, etc.) and instead try to find x and y individually. This oversight leads to unnecessary complex calculations and potential arithmetic errors.
1. Incorrect pairing of terms
Even when students recognize the pairing pattern, they might incorrectly pair terms. For example, pairing 2 with 3, 4 with 5, etc., instead of pairing 2 with 1, 4 with 3, etc. This fundamental misunderstanding of which terms correspond leads to wrong differences.
2. Arithmetic errors in the difference calculation
Students might correctly identify that each pair has a difference of 1 but then make basic multiplication errors, such as calculating \(50 \times 1 = 25\) or getting confused about how many pairs exist (thinking there are 25 pairs instead of 50).
1. Calculating \(y - x\) instead of \(x - y\)
Students might correctly find that the absolute difference is 50 but then select -50 if available in the choices, or become confused about the sign. They calculate \((1+3+5+\ldots+99) - (2+4+6+\ldots+100)\) instead of the required \(x - y\), leading them to think the answer is -50 rather than +50.
Instead of working with 50 terms directly, let's use smaller numbers to identify the pattern, then scale up.
Step 1: Start with the first 3 terms to identify the pattern
First 3 positive even integers: \(2, 4, 6\)
First 3 positive odd integers: \(1, 3, 5\)
Sum of evens: \(2 + 4 + 6 = 12\)
Sum of odds: \(1 + 3 + 5 = 9\)
Difference: \(12 - 9 = 3\)
Step 2: Check the pattern with 4 terms
First 4 positive even integers: \(2, 4, 6, 8\)
First 4 positive odd integers: \(1, 3, 5, 7\)
Sum of evens: \(2 + 4 + 6 + 8 = 20\)
Sum of odds: \(1 + 3 + 5 + 7 = 16\)
Difference: \(20 - 16 = 4\)
Step 3: Verify the pattern with 5 terms
First 5 positive even integers: \(2, 4, 6, 8, 10\)
First 5 positive odd integers: \(1, 3, 5, 7, 9\)
Sum of evens: \(2 + 4 + 6 + 8 + 10 = 30\)
Sum of odds: \(1 + 3 + 5 + 7 + 9 = 25\)
Difference: \(30 - 25 = 5\)
Step 4: Recognize the clear pattern
The pattern shows that when we take the first n positive even integers minus the first n positive odd integers, the difference equals n.
Step 5: Apply the pattern to our problem
For the first 50 positive even integers minus the first 50 positive odd integers:
Difference = 50
Why this pattern works: Each even integer is exactly 1 more than its corresponding odd integer \((2-1=1, 4-3=1, 6-5=1, \text{etc.})\), so when we have n pairs, the total difference is \(n \times 1 = n\).