已知数列｛an｝的相连两项an,an+1是方程x^2+3nx+bn=0的两个根.若a10=-17,求b51

已知数列｛an｝的相连两项an,an+1是方程x^2+3nx+bn=0的两个根.若a10=-17,求b51
已知数列｛an｝的相连两项an,an+1是方程x^2+3nx+bn=0的两个根.若a10=-17,求b51

已知数列｛an｝的相连两项an,an+1是方程x^2+3nx+bn=0的两个根.若a10=-17,求b51
an,an+1是方程x2+3nx+bn=0的两根
=>an*an+1=bn,an+an+1=-3n;
即b51=a52*a51;
an+an+1=-3n=>(an+1)+3/2(n+1)-3/4=-(an+(3/2)n-3/4);
so：[（an+1)+3/2(n+1)-3/4]/(an+(3/2)n-3/4)=-1
so:{an+(3/2)n-3/4}是公比为-1的等比数列；
so:
a52+(3/2)*52-3/4=[a10+(3/2)*10-3/4]*(-1)^(52-10);
a51+(3/2)*51-3/4=[a10+(3/2)*10-3/4]*(-1)^(51-10);
解得a52,a51带入b51=a52*a51即可解的

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...

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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

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