想到初等方法一枚：
a[n+2]=a[n+1]+a[n]
设常数k，k满足：
a[n+2]+k*a[n+1]=k*a[n+1]+a[n+1]+a[n]
a[n+2]+k*a[n+1]=(k+1)*{a[n+1]+a[n]*1/(k+1)}
计算k=1/(k+1)，得解k1与k2
于是分别有： a[n+2]+k1*a[n+1]=(k1+1)*(a[n+1]+k1*a[n])及 a[n+2]+k2*a[n+1]=(k2+1)*(a[n+1]+k2*a[n])
易知a[n+1]+k1*a[n]及a[n+1]+k2*a[n]分别为等比数列，解得该两项数列的通项公式
两数列相消即可求得a[n]的通项公式
