数列{an}中相邻两项an.an+1是方程x^2+3nx+bn=0的两根,已知a10=—17,求b51?

a(n)*a(n+1) = b(n).
a(n) + a(n+1) = -3n, -3*10=a(10) + a(11)=-17+a(11), a(11)=-13.
a(n-1)+a(n) = -3(n-1),
a(n+1)-a(n-1)=-3,

a(51) - a(49) = -3,
a(49) - a(47) = -3,
...
a(13) - a(11) = -3,

a(51) - a(11) = -3[(49-11)/2+1] = -3*20 = -60, a(51)=-60+a(11)=-60-13=-73.

a(52) - a(50) = -3,
a(50) - a(48) = -3,
...
a(12) - a(10) = -3,

a(52) - a(10) = -3[(50-10)/2+1] = -3*21 = -63, a(52)=-63+a(10)=-63-17=-80.
b(51)=a(51)*a(52)=73*80=5840

an+a(n+1)=-3n,a10=-17
a(n+2)-an=-3→a(2n)=-2-3n,a(2n+1)=2-3n
bn=an×an+1

b51=a51×a52=5840

伟达定理 an*a（n+1）=bn ....(1)
an+a（n+1）=-3n ....(2)
a10=-17 代入(2) a11=-13
同理：a11=-13 a12=-20 a13=-16 a14=-23
a15=-19
发现：an奇数行和偶数行分别成等差数列
公差d=-3 a51=a10+30d=-107代入（2）
a52=-46
b51=a50*a51= 4922