Question

已知正数数列{an}的前n项和满足Sn>1,且6Sn

Answer 1

yin wei 6Sn=(an+1)(an+2),suo yi 6Sn-1=(an-1 +1)(an-1 +2) 将两个式子相减 则可知an-an-1 =3

suo yi 等差数列 因为 6Sn=(an+1)(an+2),可知首项＝a1=2 suo yi an=3n-1

因为an((2^bn)-1)=1 an=3n-1 所以bn=log2 3n/(3n-1)=log2 3n-log2 (3n-1)

之后累加就可以求出来