a1=1
a2=1
a[n+2]=a[n+1]+a[n]
这是斐波那契数列,可以根据其特征方程x^2=x+1来求其通项
(具体的求法网上很多,
an=(1/√5)*{[(1+√5)/2]^n - [(1-√5)/2]^n}
a[2010]=(1/√5)*{[(1+√5)/2]^2010 - [(1-√5)/2]^2010}