For any positive integer n, the sum of the first n positive integers equals \(\mathrm{n(n+1)/2}\). What is the sum of...

1. Translate the problem requirements: We need to find all even integers that fall between 99 and 301 (not including 99 and 301), then calculate their sum. The given formula \(\mathrm{\frac{n(n+1)}{2}}\) gives us the sum of first n positive integers.
2. Identify the even integers in our range: Determine the first and last even integers between 99 and 301, recognizing this forms an arithmetic sequence.
3. Transform to consecutive integer problem: Use the relationship that consecutive even integers can be expressed as 2 times consecutive integers, allowing us to apply the given formula.
4. Apply the sum formula strategically: Use the given formula \(\mathrm{\frac{n(n+1)}{2}}\) to calculate the sum by finding how many consecutive integers we're dealing with.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to do in plain English. We want to find all the even numbers that sit between 99 and 301, but not including 99 and 301 themselves. Think of it like finding all the even house numbers on a street that runs from house 99 to house 301.

Since 99 is odd, the first even number after 99 is 100. Since 301 is odd, the last even number before 301 is 300. So we're looking for: 100, 102, 104, 106, ..., 296, 298, 300.

The problem also gives us a helpful formula: for any positive integer n, the sum \(\mathrm{1 + 2 + 3 + ... + n}\) equals \(\mathrm{\frac{n(n+1)}{2}}\). We'll use this formula strategically.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding of what we need to find

2. Identify the even integers in our range

Now let's identify our sequence more clearly. We have:

• First even integer: 100
• Last even integer: 300
• Pattern: each even integer is 2 more than the previous one

This forms what's called an arithmetic sequence. To find how many even integers we have, let's think about it step by step. From 100 to 300, going up by 2 each time:

• From 100 to 300, the difference is \(\mathrm{300 - 100 = 200}\)
• Since we go up by 2 each step, the number of steps is \(\mathrm{200 ÷ 2 = 100}\)
• But we need to add 1 to include both endpoints: \(\mathrm{100 + 1 = 101}\) even integers

So we have 101 even integers in our sequence.

3. Transform to consecutive integer problem

Here's the key insight: we can rewrite our even integers in a very useful way. Notice that:

• \(\mathrm{100 = 2 × 50}\)
• \(\mathrm{102 = 2 × 51}\)
• \(\mathrm{104 = 2 × 52}\)
• ...
• \(\mathrm{298 = 2 × 149}\)
• \(\mathrm{300 = 2 × 150}\)

So our sum becomes: \(\mathrm{2×50 + 2×51 + 2×52 + ... + 2×149 + 2×150}\)

We can factor out the 2: \(\mathrm{2 × (50 + 51 + 52 + ... + 149 + 150)}\)

Now we have the sum of consecutive integers from 50 to 150! This is exactly the type of problem our given formula can help us solve.

Process Skill: SIMPLIFY - Transforming a complex sequence into a simpler form we can work with

4. Apply the sum formula strategically

We need to find: \(\mathrm{50 + 51 + 52 + ... + 149 + 150}\)

The given formula tells us that \(\mathrm{1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}}\)

To use this formula, we'll calculate:

• Sum from 1 to 150: \(\mathrm{150 × 151 ÷ 2 = 11,325}\)
• Sum from 1 to 49: \(\mathrm{49 × 50 ÷ 2 = 1,225}\)
• Sum from 50 to 150: \(\mathrm{11,325 - 1,225 = 10,100}\)

Therefore, our final answer is: \(\mathrm{2 × 10,100 = 20,200}\)

Let's verify: We have 101 even integers, with an average value of \(\mathrm{(100 + 300) ÷ 2 = 200}\). So \(\mathrm{101 × 200 = 20,200}\) ✓

4. Final Answer

The sum of all even integers between 99 and 301 is 20,200.

This matches answer choice B. 20,200.

Common Faltering Points

Errors while devising the approach

Students often include 99 and 301 when finding integers "between" these numbers. Since 99 and 301 are both odd, this wouldn't affect the even integer count, but if the endpoints were even, this misinterpretation would lead to including extra terms in the sum.

Many students attempt to directly sum the arithmetic sequence \(\mathrm{100 + 102 + 104 + ... + 300}\) without realizing they can factor out 2 to get \(\mathrm{2×(50 + 51 + 52 + ... + 150)}\). This makes the problem much more complex and prone to computational errors.

Students frequently forget to add 1 when counting terms in a sequence. For even integers from 100 to 300, they might calculate \(\mathrm{(300-100)/2 = 100}\) terms instead of the correct 101 terms, leading to an incorrect application of formulas.

Errors while executing the approach

When calculating Sum(1 to 150) - Sum(1 to 49), students often make computational mistakes: \(\mathrm{150×151/2 = 11,325}\) or \(\mathrm{49×50/2 = 1,225}\). These calculations involve large numbers and are error-prone under time pressure.

After correctly finding that \(\mathrm{50 + 51 + ... + 150 = 10,100}\), students sometimes forget that this represents the sum of (50, 51, 52, ..., 150) and they need to multiply by 2 to get back to the original even integers (100, 102, 104, ..., 300).

Students might incorrectly calculate Sum(1 to 49) instead of understanding they need Sum(1 to 49) to subtract from Sum(1 to 150), or they might use Sum(1 to 50) instead, leading to an off-by-one error in their final calculation.

Errors while selecting the answer

No likely faltering points - the calculation directly yields 20,200 which matches answer choice B exactly, with built-in verification through the average method.