The sum of the first k positive integers is equal to \(\frac{\mathrm{k}(\mathrm{k}+1)}{2}\). What is the sum of the integers from...

1. Translate the problem requirements: We need to find the sum of consecutive integers from n to m (inclusive), where both n and m are positive and \(\mathrm{n < m}\). The given formula calculates sum from 1 to k, but we need sum from n to m.
2. Recognize the sum relationship pattern: Since we have a formula for sum from 1 to any number, we can use subtraction to isolate the sum from n to m by taking the total sum up to m and removing the unwanted portion.
3. Identify what to subtract: To get sum from n to m, we need sum from 1 to m minus sum from 1 to (n-1), since this removes all integers from 1 to n-1, leaving only n to m.
4. Apply the given formula correctly: Use the formula with k=m for the total sum, and k=(n-1) for the portion to subtract, then match with answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to find. We need the sum of consecutive integers from n to m, where both numbers are positive and n is less than m.

For example, if n = 3 and m = 7, we want: \(\mathrm{3 + 4 + 5 + 6 + 7}\)

The problem gives us a formula for adding up the first k positive integers (like \(\mathrm{1 + 2 + 3 + ... + k}\)), which equals \(\mathrm{k(k+1)/2}\). But we don't want to start from 1 - we want to start from some number n and go up to m.

Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding

2. Recognize the sum relationship pattern

Here's the key insight: we can use the given formula creatively. Think of it like this - if you want to find how much money you spent from Tuesday to Friday, and you know your total spending from Monday to Friday, you just subtract what you spent on Monday.

Similarly, to get the sum from n to m, we can:

• Calculate the sum from 1 to m (using our formula)
• Subtract the sum from 1 to some number (to remove the unwanted part)
• This leaves us with just the sum from n to m

For our example (n=3, m=7):

• Sum from 1 to 7 = \(\mathrm{1+2+3+4+5+6+7 = 28}\)
• Sum from 1 to 2 = \(\mathrm{1+2 = 3}\)
• Sum from 3 to 7 = \(\mathrm{28 - 3 = 25}\)

3. Identify what to subtract

To get the sum from n to m, we need to remove all the numbers before n. That means we subtract the sum from 1 to (n-1).

Why (n-1)? Because if we want to start our sum at n, we need to remove everything from 1 up to the number just before n, which is (n-1).

In our example: to start at 3, we remove 1 and 2, so we subtract the sum from \(\mathrm{1\ to\ (3-1) = 1\ to\ 2}\).

Process Skill: INFER - Drawing the non-obvious conclusion about what exactly needs to be subtracted

4. Apply the given formula correctly

Now we can write this mathematically:

Sum from n to m = (Sum from 1 to m) - (Sum from 1 to n-1)

Using our formula \(\mathrm{k(k+1)/2}\):

• Sum from 1 to m = \(\mathrm{m(m+1)/2}\)
• Sum from 1 to (n-1) = \(\mathrm{(n-1)n/2}\)

Therefore: Sum from n to m = \(\mathrm{m(m+1)/2 - (n-1)n/2}\)

Looking at the answer choices, this matches option C: \(\mathrm{m(m+1)/2 - (n-1)n/2}\)

5. Final Answer

The sum of integers from n to m, inclusive, is \(\mathrm{m(m+1)/2 - (n-1)n/2}\).

This corresponds to answer choice C.

To verify: this formula takes the complete sum up to m and subtracts exactly the portion from 1 to n-1, leaving us with the sum from n to m as required.

Common Faltering Points

Errors while devising the approach

• Misunderstanding the range of summation: Students often confuse whether the sum should be "from n to m" (starting at n) or "from 1 to m minus 1 to n" and may incorrectly think they need to subtract the sum from 1 to n instead of 1 to (n-1). This leads to wrong setup of the subtraction formula.
• Incorrectly applying the given formula: Students may try to directly modify the given formula \(\mathrm{k(k+1)/2}\) by substituting different values without recognizing that this formula specifically calculates sums starting from 1, not from an arbitrary starting point n.
• Overlooking the "inclusive" requirement: Students might miss that the sum needs to include both endpoints n and m, leading them to incorrectly exclude one of these boundary values in their approach.

Errors while executing the approach

• Arithmetic errors in the subtraction setup: When setting up Sum from 1 to m minus Sum from 1 to (n-1), students may incorrectly write this as \(\mathrm{m(m+1)/2 - n(n+1)/2}\) instead of the correct \(\mathrm{m(m+1)/2 - (n-1)n/2}\), essentially using n instead of (n-1).
• Expanding the wrong terms: Students may incorrectly expand (n-1)n as (n-1)(n-1) or confuse the order, writing it as n(n-1) instead of (n-1)n, leading to computational errors.

Errors while selecting the answer

• Confusing similar-looking answer choices: Students may correctly derive \(\mathrm{m(m+1)/2 - (n-1)n/2}\) but then select option B which shows \(\mathrm{m(m+1)/2 - n(n+1)/2}\), mistaking the similar structure and not carefully checking the second term.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose convenient values for n and m

Let's select n = 3 and m = 6. These are small positive integers that satisfy \(\mathrm{0 < n < m}\), making calculations manageable while testing our formula.

Step 2: Calculate the actual sum from 3 to 6 directly

Sum from 3 to 6 = \(\mathrm{3 + 4 + 5 + 6 = 18}\)

Step 3: Test each answer choice with our smart numbers

Using the given formula for sum of first k positive integers = \(\mathrm{k(k+1)/2}\):

Choice A: \(\mathrm{m(m+1)/2 - (n+1)(n+2)/2 = 6(7)/2 - 4(5)/2 = 21 - 10 = 11}\) ❌

Choice B: \(\mathrm{m(m+1)/2 - n(n+1)/2 = 6(7)/2 - 3(4)/2 = 21 - 6 = 15}\) ❌

Choice C: \(\mathrm{m(m+1)/2 - (n-1)n/2 = 6(7)/2 - 2(3)/2 = 21 - 3 = 18}\) ✓

Choice D: \(\mathrm{(m-1)m/2 - (n+1)(n+2)/2 = 5(6)/2 - 4(5)/2 = 15 - 10 = 5}\) ❌

Choice E: \(\mathrm{(m-1)m/2 - n(n+1)/2 = 5(6)/2 - 3(4)/2 = 15 - 6 = 9}\) ❌

Step 4: Verify with different smart numbers

Let's confirm with n = 2, m = 5:

Direct sum: \(\mathrm{2 + 3 + 4 + 5 = 14}\)

Choice C: \(\mathrm{5(6)/2 - 1(2)/2 = 15 - 1 = 14}\) ✓

Conclusion: Choice C consistently gives the correct result, confirming our answer.