Question

AL95
Cum ar trebui să procedez cu exerciții de genul?

Answer 1

$M = \sqrt{C_{2n+1}^1 \cdot C_{2n+1}^2\cdot C_{2n+1}^3\cdot ... \cdot C_{2n+1}^{2n}},~n\in \mathbb{N}^* \\ \\ \text{Not\u{a}m }2n+1 = m \Rightarrow 2n = m-1 \\ n\in \mathbb{N}^* \Rightarrow 2n+1 \in \mathbb{N}^* \Rightarrow m \in \mathbb{N}^* \\ \\ \Rightarrow M = \sqrt{C_{m}^{1}\cdot C_{m}^2\cdot C_{m}^3\cdot ...\cdot C_{m}^{m-1}},~m\in \mathbb{N}^*\\ \\ \Rightarrow M = \sqrt{\Big(C_{m}^1\cdot C_{m}^{m-1}\Big)\cdot\Big(C_{m}^2\cdot C_{m}^{m-2}\Big)\cdot....\cdot??? }$

$2n \rightarrow \text{par } \Rightarrow m-1 \rightarrow \text{par} \Rightarrow m \rightarrow \text{impar}$

$\text{Lu\u{a}m ca exemplu un num\u{a}r impar:}\\ \\ m = 5 \\ \\ \Rightarrow M = \sqrt{\Big(C_{5}^1\cdot C_{5}^4\Big)\cdot \Big(C_{5}^{2}\cdot C_{5}^3\Big)} = \sqrt{\Big(C_{5}^1\cdot C_{5}^4\Big)\cdot \Big(C_{5}^{2}\cdot C_{5}^{3}\Big)}= \\ \\ = \sqrt{\Big(C_{5}^1\cdot C_{5}^4\Big)\cdot \Big(C_{5}^{\frac{5-1}{2}}\cdot C_{5}^{\frac{5-1}{2}+1}\Big)} = \\ \\ = \sqrt{\Big(C_{5}^1\Big)^2\cdot \Big(C_{5}^2\Big)^2 } = \sqrt{5^2\cdot 10^2} = 50 \in \mathbb_{N}^*$

$\text{Analog:}\\ \\ \Rightarrow M = \sqrt{\Big(C_{m}^{1}\Big)^2 \cdot \Big(C_{m}^{2}\Big)^2\cdot ... \cdot \Big(C_{m}^{\frac{m-1}{2}}\Big)^2} \in \mathbb{N}^* \Rightarrow \boxed{\text{d) corect}}$