若a²=a+1,b²=b+1,a≠b,则a^5+b^5等于多少?

若a²=a+1,b²=b+1,a≠b,则a^5+b^5等于多少?
若a²=a+1,b²=b+1,a≠b,则a^5+b^5等于多少?

若a²=a+1,b²=b+1,a≠b,则a^5+b^5等于多少?
实际上a和b就是方程x^2=x+1（即x^2-x-1=0）的2个根.
根据求根公式,
x=[1±√(1+4)]/2
=(1±√5)/2
所以设a大于b（其实a小于b也一样）
则：a=(1+√5)/2
b=(1-√5)/2
a^5=a^2*a^2*a=(a+1)^2*a=(3+√5)^2/4*(1+√5)/2=(44+20√5)/8
b^5=b^2*b^2*b=(b+1)^2*a=(3-√5)^2/4*(1-√5)/2=(44-20√5)/8
所以a^5+b^5 =(44+20√5)/8+(44-20√5)/8=88/8=11

a^5+b^5
=(a²)(a²)(a)+(b²)(b²)(b)
=(a+1)²*a+(b+1)²*b
=a*(a²+2a+1)+b*(b²+2b+1)
=a*(a+1+2a+1)+b*(b+1+2b+1)
=a*(3a+2)+b*(3b+2)
=3a²+2a+3b²+2b
=3a+3+2a+3b+3+2b
=5a+5b+6