Question
\( n \) is an integer. Prove algebraically that the sum of \( \frac{1}{2} n(n+1) \) and \( \frac{1}{2}(n+1)(n+2) \) is always a square number.
Ask by Craig Cross. in the United Kingdom
Jan 23,2025
Real Tutor Solution
Tutor-Verified Answer
Answer
The sum simplifies to \( (n+1)^2 \), which is always a square number.
Solution
To prove that the sum of \( \frac{1}{2} n(n+1) \) and \( \frac{1}{2}(n+1)(n+2) \) is always a square number, let's perform the algebraic computation step by step.
**Given:**
\[
S = \frac{1}{2} n(n+1) + \frac{1}{2}(n+1)(n+2)
\]
**Step 1: Factor out \(\frac{1}{2}\) and \((n+1)\)**
\[
S = \frac{1}{2} \left[ n(n+1) + (n+1)(n+2) \right]
\]
\[
= \frac{1}{2} (n+1) \left[ n + (n+2) \right]
\]
**Step 2: Simplify the expression inside the brackets**
\[
n + (n+2) = 2n + 2
\]
**Step 3: Substitute back into the expression**
\[
S = \frac{1}{2} (n+1) (2n + 2)
\]
**Step 4: Factor out the 2 from \(2n + 2\)**
\[
S = \frac{1}{2} \times 2 \times (n+1)(n+1)
\]
**Step 5: Simplify the coefficients**
\[
S = (n+1)^2
\]
**Conclusion:**
\[
S = (n+1)^2
\]
Thus, the sum \( \frac{1}{2} n(n+1) + \frac{1}{2}(n+1)(n+2) \) simplifies to \( (n+1)^2 \), which is clearly a perfect square for any integer \( n \).
**Final Answer:**
After simplifying, the sum is (n + 1)². Thus, the sum is always a square number.
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution The Deep Dive
To prove that the sum \( \frac{1}{2} n(n+1) + \frac{1}{2} (n+1)(n+2) \) is always a square number, let's first simplify the expression. Starting with the sum: \[ \frac{1}{2} n(n+1) + \frac{1}{2} (n+1)(n+2) \] Factoring out \( \frac{1}{2} (n+1) \): \[ = \frac{1}{2} (n+1) \left( n + (n+2) \right) \] \[ = \frac{1}{2} (n+1) (2n + 2) \] \[ = \frac{1}{2} (n+1) \cdot 2(n + 1) \] \[ = (n+1)^2 \] Now, observing that \( (n+1)^2 \) is a perfect square for any integer \( n \). Thus, we have shown algebraically that the sum \( \frac{1}{2} n(n+1) + \frac{1}{2} (n+1)(n+2) \) is always a square number.
