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The sum of the first in positive integers is given by \(\mathrm{n(n + 1)/2}\). What is the sum of the first 100 positive odd integers?
Let's start by understanding exactly what we're looking for. The problem asks for the sum of the first 100 positive odd integers. These are the numbers: 1, 3, 5, 7, 9, 11, 13, 15, ... and we want to keep going until we've added up exactly 100 of these odd numbers.
Notice that the formula given in the problem, \(\mathrm{n(n+1)/2}\), is for adding up consecutive positive integers starting from 1 (like \(\mathrm{1+2+3+4+...}\)). But we only want the odd ones, so this formula doesn't directly apply to our situation.
Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding
Let's look at the pattern of positive odd integers more carefully:
Do you see the pattern? Each odd integer is 2 more than the previous one. If we want to find the nth odd integer, we can use the pattern: the nth odd integer equals \(\mathrm{(2n - 1)}\).
For example:
So the 100th odd integer would be \(\mathrm{2(100) - 1 = 199}\).
Here's where we can use a beautiful mathematical insight! Let's look at what happens when we add up the first few odd integers:
Wow! Do you see the pattern? The sum of the first n positive odd integers always equals \(\mathrm{n^2}\)!
This makes intuitive sense if you think about it visually - odd integers can be arranged in perfect squares. This is a fundamental property that always works, no matter how large n gets.
Process Skill: INFER - Drawing the non-obvious conclusion that the sum equals \(\mathrm{n^2}\)
Now we can apply our discovered pattern directly. Since we want the sum of the first 100 positive odd integers, and we know that the sum of the first n odd integers equals \(\mathrm{n^2}\), our answer is simply:
\(\mathrm{100^2 = 100 \times 100 = 10,000}\)
Let's verify this makes sense: we're adding 100 different odd numbers, starting from small ones like 1, 3, 5 and going up to larger ones near 199. An average value might be around 100, so \(\mathrm{100 \times 100 = 10,000}\) seems very reasonable.
The sum of the first 100 positive odd integers is 10,000.
Looking at our answer choices, this matches choice B exactly.
Answer: B. 10,000
Students often see the formula \(\mathrm{n(n+1)/2}\) in the problem and immediately try to substitute \(\mathrm{n=100}\), getting \(\mathrm{100(101)/2 = 5,050}\). They miss that this formula applies to consecutive integers (1,2,3,4...) but the problem asks for odd integers only (1,3,5,7...).
Students may misinterpret "first 100 positive odd integers" and instead try to find the sum of odd integers from 1 to 100. This would mean summing \(\mathrm{1+3+5+...+99}\), which gives only 50 odd numbers instead of 100.
Students may struggle to establish that the nth odd integer equals \(\mathrm{(2n-1)}\). Without this pattern, they cannot identify that the 100th odd integer is 199, making it difficult to verify their final answer or use alternative solution methods.
When checking the pattern that sum of first n odd integers equals \(\mathrm{n^2}\), students may make calculation errors. For example, computing \(\mathrm{1+3+5+7 = 15}\) instead of 16, leading them to doubt the \(\mathrm{n^2}\) pattern and abandon the correct approach.
Even after correctly identifying that the answer should be \(\mathrm{100^2}\), students may make computational errors like \(\mathrm{100^2 = 1,000}\) instead of 10,000, especially under time pressure.
Students who calculated both \(\mathrm{100(101)/2 = 5,050}\) and \(\mathrm{100^2 = 10,000}\) during their work might select answer choice E (5,050) instead of B (10,000), confusing which calculation applies to the actual question being asked.