$$ n $$ 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.

Question

UpStudy AI · Accepted Answer

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.
The sum simplifies to $$ (n+1)^2 $$, which is always a square number.

\( n \) 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.

Solución de tutoría real

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Algebra Questions