根據(jù)韋達定理,因為an與an+1是方程兩個解
所以a(n)+a(n+1)=-b/a=-3n/1=-3n
a(n)+a(n+1)=-3n,
a(n+1)=-a(n)-3n=-a(n)-3n/2 - 3(n+1)/2 +3/2
a(n+1)+3(n+1)/2 = -[a(n)+3n/2] + 3/2 = -[a(n)+3n/2] + 3/4 + 3/4,
a(n+1)+3(n+1)/2 - 3/4 = -[a(n)+3n/2-3/4],
{a(n)+3n/2-3/4}是首項為a(1)+3/2-3/4=2+3/4=11/4,公比為（-1）的等比數(shù)列.
a(n)+3n/2-3/4 = (11/4)(-1)^(n-1),
所以通項公式是a(n) = 3/4 - 3n/2 + (11/4)(-1)^(n-1),
有根據(jù)韋達定理a(n)*a(n+1)=c/a=Cn
c(100)=a(100)*a(101)=[3/4-300/2+(11/4)(-1)^(99)][3/4-303/2+(11/4)(-1)^(100)]
=[3/4-300/2-11/4][3/4-303/2+11/4]
=[-608/4][-592/4]
=[152][148]
=[150+2][150-2]
=(150)^2-4
=22500-4
=22496