等比数列{an}是等差数列,公差d>0,Sn是{an}的前n项和,已知a2a3=40,S4+26,1.求数列{an}的通项公式an.2.令bn=1/ana(n+1),求数列{bn}的所有项之和T

Question

等比数列{an}是等差数列,公差d>0,Sn是{an}的前n项和,已知a2a3=40,S4+26,
1.求数列{an}的通项公式an.2.令bn=1/ana(n+1),求数列{bn}的所有项之和T

Answer 1

S4=(a1+a4)×4/2=26
a1+a4=13
则 a2+a3=13
又已知a2a3=40
解得 a2=5 a3=8
则 a1=2 a4=11
则 an=3n-1
bn=1/ana(n+1)=1/(3n-1)(3n+1)
T=1/(2×5)+1/(5×8)+.+1/(3n-1)(3n+2)
=(1/2-1/5+1/5-1/8+,+1/(3n-1)-1/(3n+2))/3
=(1/2-1/(3n+2))/3