If the sum of the first n positive integers is S, what is the sum of the first n positive...

1. Translate the problem requirements: We need to understand what S represents (sum of first n positive integers: \(1+2+3+\ldots+\mathrm{n}\)) and find an expression for the sum of first n positive even integers (\(2+4+6+\ldots+2\mathrm{n}\)) in terms of S
2. Identify the relationship between sequences: Compare the first n positive integers with the first n positive even integers to spot the pattern
3. Apply factoring to establish the connection: Use the relationship discovered to express the even integer sum in terms of the original sum S
4. Verify with concrete example: Test the relationship with specific values to confirm our mathematical reasoning

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're dealing with in plain English.

We're told that S represents the sum of the first n positive integers. So if we take the first 4 positive integers, that would be: \(1 + 2 + 3 + 4 = 10\). In this case, \(\mathrm{S} = 10\).

Now we need to find the sum of the first n positive even integers. For the first 4 positive even integers, that would be: \(2 + 4 + 6 + 8\).

Our job is to express this sum of even integers in terms of S (which we know is 10 in our example).

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

2. Identify the relationship between sequences

Let's look at both sequences side by side to spot a pattern:

First n positive integers: 1, 2, 3, 4, 5, 6, ...

First n positive even integers: 2, 4, 6, 8, 10, 12, ...

Notice something interesting? Each even integer is exactly double the corresponding positive integer in the same position!

• The 1st even integer (2) = 2 × 1st positive integer (1)

• The 2nd even integer (4) = 2 × 2nd positive integer (2)

• The 3rd even integer (6) = 2 × 3rd positive integer (3)

This pattern continues: every even integer is exactly twice the corresponding positive integer.

3. Apply factoring to establish the connection

Since each even integer is double the corresponding positive integer, what happens when we add them all up?

Sum of first n even integers = \(2 + 4 + 6 + \ldots + 2\mathrm{n}\)

We can factor out the 2 from each term:

= \(2(1) + 2(2) + 2(3) + \ldots + 2(\mathrm{n})\)

= \(2(1 + 2 + 3 + \ldots + \mathrm{n})\)

But wait! The expression \((1 + 2 + 3 + \ldots + \mathrm{n})\) is exactly what S represents!

Therefore: Sum of first n even integers = \(2\mathrm{S}\)

4. Verify with concrete example

Let's test our reasoning with n = 4:

First 4 positive integers: \(1 + 2 + 3 + 4 = 10\)

So \(\mathrm{S} = 10\)

First 4 positive even integers: \(2 + 4 + 6 + 8 = 20\)

According to our formula: \(2\mathrm{S} = 2(10) = 20\) ✓

Perfect! Our relationship checks out.

Final Answer

The sum of the first n positive even integers equals \(2\mathrm{S}\).

The correct answer is (C) \(2\mathrm{S}\).

Common Faltering Points

Errors while devising the approach

• Misinterpreting "first n positive even integers": Students often confuse this with "the first n even positions in the sequence of positive integers." For example, when n=4, they might think we want the even numbers at positions 1,2,3,4 of all positive integers (which would be 2,4) rather than the first 4 even numbers (2,4,6,8).
• Not recognizing the doubling pattern: Students may try to derive complex formulas or use arithmetic series formulas instead of noticing the simple relationship that each even integer is exactly twice the corresponding positive integer in the same position.
• Attempting to solve using formulas rather than relationships: Students might try to use the arithmetic series formula \(\mathrm{S} = \mathrm{n}(\mathrm{n}+1)/2\) for both sequences and then find their ratio, making the problem unnecessarily complex instead of using the direct factoring approach.

Errors while executing the approach

• Incorrect factoring: When factoring out 2 from the sum \(2+4+6+\ldots+2\mathrm{n}\), students might write it as \(2(2+4+6+\ldots)\) instead of the correct \(2(1+2+3+\ldots)\), failing to properly extract the factor from each term.
• Arithmetic errors in verification: During the concrete example check with n=4, students may incorrectly calculate either \(\mathrm{S}=1+2+3+4\) or the sum \(2+4+6+8\), leading them to doubt their correct approach.

Errors while selecting the answer

• Confusion between S and 2S: Even after correctly deriving that the sum equals \(2\mathrm{S}\), students might second-guess themselves and select S (choice B), thinking "the relationships should be simpler" or misremembering their derivation.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for n
Let's use n = 3, which gives us a manageable set of numbers to work with.

Step 2: Calculate S (sum of first 3 positive integers)
First 3 positive integers: 1, 2, 3
\(\mathrm{S} = 1 + 2 + 3 = 6\)

Step 3: Find the sum of first 3 positive even integers
First 3 positive even integers: 2, 4, 6
\(\mathrm{Sum} = 2 + 4 + 6 = 12\)

Step 4: Express the relationship in terms of S
We found that when \(\mathrm{S} = 6\), the sum of even integers = 12
Notice that \(12 = 2 \times 6 = 2\mathrm{S}\)

Step 5: Verify with another value
Let's try n = 4:
First 4 positive integers: 1, 2, 3, 4 → \(\mathrm{S} = 10\)
First 4 positive even integers: 2, 4, 6, 8 → \(\mathrm{Sum} = 20\)
Again, \(20 = 2 \times 10 = 2\mathrm{S}\)

Step 6: Check answer choices
Our pattern shows the sum of first n positive even integers = \(2\mathrm{S}\)
This matches choice (C) \(2\mathrm{S}\).

Why this works: Each positive even integer is exactly double its corresponding positive integer in the same position (1st even = 2×1st positive, 2nd even = 2×2nd positive, etc.), so the total sum is doubled.