x^2-12x+27=0
(x-3)(x-9)=0;
x1=3,x3=9;
a5>a2,所以a2=3,a5=9
d=(a5-a2)/3=2;
a1=a2-d=1;
an=1+2(n-1)=2n-1
sn,bn,cn又是啥?
a2,a5是方程x^2-12x+27=0的两根,数列{an}是递增的等差数列,数列{bn}的前n项和为Sn记cn=anbn,求sn
a2,a5是方程x^2-12x+27=0的两根,数列{an}是递增的等差数列,数列{bn}的前n项和为Sn记cn=anbn,求sn
数学人气：517 ℃时间：2020-02-06 03:21:24
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