- Translate the problem requirements: We need to find an equivalent expression for \(\mathrm{n! + (n + 1)! + (n + 2)!}\) by understanding what each factorial means and how they relate to each other.
- Express each factorial in terms of n!: Rewrite \(\mathrm{(n + 1)!}\) and \(\mathrm{(n + 2)!}\) using the fundamental property that each factorial builds on the previous one.
- Factor out the common term: Extract \(\mathrm{n!}\) as a common factor from all three terms to simplify the expression.
- Simplify the remaining expression: Combine the terms inside the parentheses and match the result with the given answer choices.
Execution of Strategic Approach
1. Translate the problem requirements
We need to find what \(\mathrm{n! + (n + 1)! + (n + 2)!}\) equals when simplified. Let's first understand what each factorial means in everyday terms.
Think of factorials as expanding patterns:
- \(\mathrm{n!}\) means "multiply all whole numbers from 1 up to n"
- \(\mathrm{(n + 1)!}\) means "multiply all whole numbers from 1 up to (n + 1)"
- \(\mathrm{(n + 2)!}\) means "multiply all whole numbers from 1 up to (n + 2)"
The key insight is that each larger factorial contains all the previous factorials as building blocks.
Process Skill: TRANSLATE - Converting the factorial notation into conceptual understanding
2. Express each factorial in terms of n!
Let's see how each factorial relates to \(\mathrm{n!}\) by thinking about what happens when we expand them:
For \(\mathrm{(n + 1)!}\): This means we multiply all numbers from 1 to (n + 1). But we can think of this as "first multiply 1 through n (which gives us \(\mathrm{n!}\)), then multiply by the next number (n + 1)". So:
\(\mathrm{(n + 1)! = (n + 1) \times n!}\)
For \(\mathrm{(n + 2)!}\): Similarly, this means we multiply all numbers from 1 to (n + 2). We can build this step by step: "first get \(\mathrm{(n + 1)!}\), then multiply by (n + 2)". So:
\(\mathrm{(n + 2)! = (n + 2) \times (n + 1)! = (n + 2) \times (n + 1) \times n!}\)
Now we can rewrite our original expression:
\(\mathrm{n! + (n + 1)! + (n + 2)! = n! + (n + 1) \times n! + (n + 2)(n + 1) \times n!}\)
3. Factor out the common term
Now we can see that every term contains \(\mathrm{n!}\) as a factor. Let's pull out this common factor, just like factoring out a common number in regular algebra:
\(\mathrm{n! + (n + 1) \times n! + (n + 2)(n + 1) \times n!}\)
\(\mathrm{= n! \times [1 + (n + 1) + (n + 2)(n + 1)]}\)
This is the same principle as factoring \(\mathrm{3x + 6x + 9x = 3x(1 + 2 + 3)}\), but with factorials.
4. Simplify the remaining expression
Now let's simplify what's inside the brackets:
\(\mathrm{[1 + (n + 1) + (n + 2)(n + 1)]}\)
Let's work through this step by step:
- First term: 1
- Second term: \(\mathrm{(n + 1)}\)
- Third term: \(\mathrm{(n + 2)(n + 1) = (n + 1)(n + 2) = n^2 + 2n + n + 2 = n^2 + 3n + 2}\)
Adding them all together:
\(\mathrm{1 + (n + 1) + (n^2 + 3n + 2)}\)
\(\mathrm{= 1 + n + 1 + n^2 + 3n + 2}\)
\(\mathrm{= n^2 + 4n + 4}\)
\(\mathrm{= (n + 2)^2}\)
Therefore: \(\mathrm{n! + (n + 1)! + (n + 2)! = n! \times (n + 2)^2 = n!(n + 2)^2}\)
Final Answer
Our simplified expression is \(\mathrm{n!(n + 2)^2}\), which matches answer choice D.
Let's verify with a quick example: if \(\mathrm{n = 3}\)
- Original: \(\mathrm{3! + 4! + 5! = 6 + 24 + 120 = 150}\)
- Our answer: \(\mathrm{3! \times (3 + 2)^2 = 6 \times 25 = 150}\) ✓
The answer is D.
Common Faltering Points
Errors while devising the approach
- Misunderstanding factorial relationships: Students may not recognize that larger factorials contain smaller factorials as building blocks. They might try to calculate each factorial separately (like computing \(\mathrm{5! = 120}\) from scratch) instead of seeing that \(\mathrm{(n+1)! = (n+1) \times n!}\). This leads to much more complex calculations and potential arithmetic errors.
- Missing the factoring opportunity: Students may not realize that they should factor out the common \(\mathrm{n!}\) term from all three expressions. Instead, they might attempt to expand everything or try to match the expression directly to the answer choices without simplification, making the problem unnecessarily difficult.
Errors while executing the approach
- Algebraic expansion errors: When expanding \(\mathrm{(n+2)(n+1)}\), students commonly make mistakes like getting \(\mathrm{n^2 + 3n + 1}\) instead of \(\mathrm{n^2 + 3n + 2}\), or forgetting to distribute properly. These small arithmetic errors cascade through the rest of the solution.
- Combining like terms incorrectly: When adding \(\mathrm{1 + (n+1) + (n^2 + 3n + 2)}\), students may incorrectly combine the constant terms (getting 3 instead of 4) or the linear terms, leading to an expression other than \(\mathrm{n^2 + 4n + 4}\).
- Factoring the final quadratic: Students may not recognize that \(\mathrm{n^2 + 4n + 4}\) is a perfect square trinomial equal to \(\mathrm{(n+2)^2}\). They might leave it in expanded form or factor it incorrectly, preventing them from matching it to answer choice D.
Errors while selecting the answer
- Verification confusion: Even if students arrive at \(\mathrm{n!(n+2)^2}\), they might second-guess themselves and choose a 'simpler-looking' option like \(\mathrm{3(n+1)!}\) or \(\mathrm{n!(n+3)}\), thinking their answer seems too complex or that they made an error somewhere in the process.
Alternate Solutions
Smart Numbers Approach
Step 1: Choose a convenient value for n
Let's use \(\mathrm{n = 3}\) because it's small enough to calculate factorials easily but large enough to see the patterns clearly.
Step 2: Calculate the factorials
- \(\mathrm{n! = 3! = 6}\)
- \(\mathrm{(n + 1)! = 4! = 24}\)
- \(\mathrm{(n + 2)! = 5! = 120}\)
Step 3: Add the factorials
\(\mathrm{n! + (n + 1)! + (n + 2)! = 6 + 24 + 120 = 150}\)
Step 4: Test each answer choice with n = 3
- Choice A: \(\mathrm{(n!)^3 = 6^3 = 216 \neq 150}\)
- Choice B: \(\mathrm{3(n + 1)! = 3(24) = 72 \neq 150}\)
- Choice C: \(\mathrm{n!(n + 1)^3 = 6(4^3) = 6(64) = 384 \neq 150}\)
- Choice D: \(\mathrm{n!(n + 2)^2 = 6(5^2) = 6(25) = 150}\) ✓
- Choice E: \(\mathrm{n!(n + 3) = 6(6) = 36 \neq 150}\)
Step 5: Verify with another value
Let's verify with \(\mathrm{n = 2}\):
- \(\mathrm{2! + 3! + 4! = 2 + 6 + 24 = 32}\)
- Choice D: \(\mathrm{n!(n + 2)^2 = 2!(4^2) = 2(16) = 32}\) ✓
Conclusion: The answer is D. \(\mathrm{n!(n + 2)^2}\)
