Question

设f(x)=1/4x^2+1/2x-3/4,正数数列{an}的前n项和为Sn,且Sn=f(an),(n属于正整数集),(

Answer 1

a1b1+a2b2+```````+anbn=（2n-1）*2^(n+1)+2

a1b1+a2b2+```````+an-1bn-1=（2n-3）*2^n+2

两式子相减得anbn=（2n-1）*2^(n+1)-（2n-3）*2^n

=4n*2^n-2*2^n-2n*2^n+3*2^n=(2n+1)*2^n又第一小问答案是2n+1即an=2n+1

则bn=2^n

计算过程不敢包,但方法一定对