题目

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=

提问时间：2021-02-21

答案

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