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Find all sequences $(a_n)_{n\geq 1}$ of positive integers such that for all integers $n\geq 3$ we have
\[\dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}= 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}.\]
求所有的正整数列$(a_n)_{n\geq 1}$,满足对一切正整数$n\geq 3$都有\[\dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n}= 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}.\]
考虑$n=3$,可得\[ \frac{1}{a_1 a_3}=1-\frac{1}{a_1^2+a_2^2},a_1 a_3=\frac{a_1^2+a_2^2}{a_1^2+a_2^2-1}\]
所以必须有$a_1^2+a_2^2-1=1,a_1=a_2=1,a_3=2$
考虑\begin{gather} \dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-2}a_n} = 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2}\\ \dfrac{1}{a_1 a_3} + \dfrac{1}{a_2a_4} + \cdots + \dfrac{1}{a_{n-1}a_{n+1}} = 1 – \dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2+a_n^2}\\ \frac{1}{a_{n-1}a_{n+1}}=\dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2} -\dfrac{1}{a_1^2+a_2^2+\cdots +a_{n-1}^2+a_n^2}\\ (a_1^2+a_2^2+\cdots +a_{n-1}^2)(a_1^2+a_2^2+\cdots +a_{n}^2)=a_{n-1} a_n^2 a_{n+1}\\ \frac{a_1^2+a_2^2+\cdots +a_{n-1}^2}{a_{n-1} a_n} \frac{a_1^2+a_2^2+\cdots +a_{n}^2}{a_{n} a_{n+1}}=1 \end{gather}
记$b_n=\frac{a_1^2+a_2^2+\cdots +a_{n}^2}{a_{n} a_{n+1}},b_{n-1}b_n=1,n \ge 3$
结合$b_1=b_2=1$,可得对一切$n \in \mN$都有$b_n=1$\begin{gather}a_1^2+a_2^2+\cdots +a_{n}^2=a_{n} a_{n+1}\\a_1^2+a_2^2+\cdots +a_{n+1}^2=a_{n+1} a_{n+2}\\a_{n+1}^2=a_{n+1} a_{n+2}-a_{n} a_{n+1}\\a_{n+2}=a_{n+1}+a_n\end{gather}
所以$\{a_n\}$是斐波那契数列,通项为\[ a_n=\frac{1}{\sqrt{5}}\left((\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n\right)\]
