若数列an中，相邻两项an,an+1是函数y=x^2+3nx+bn的两个零点，已知a10=-7,则b51=

来源：百度知道编辑：UC知道时间：2024/09/25 05:34:31

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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