a1=1
a2=1
a[n+2]=a[n+1]+a[n]
這是斐波那契數(shù)列,可以根據(jù)其特征方程x^2=x+1來求其通項
(具體的求法網(wǎng)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}