已知数列{an}满足条件:a1=1,a2=r(r>0{anan+1}是公比为q(q>0)的等比数列.设bn =a(2n-1)+a(2n)
问题描述:
已知数列{an}满足条件:a1=1,a2=r(r>0{anan+1}是公比为q(q>0)的等比数列.设bn =a(2n-1)+a(2n)
求{bn}的通项公式及{bn}的前n项和Sn
求过程谢谢
答
∵数列{a[n]}满足条件:a[1]=1,a[2]=r,且数列{a[n]a[n+1]}是公比为q的等比数列∴q≠0,r≠0,且a[n]a[n+1]=a[1]a[2]q^(n-1)=rq^(n-1)∵a[n]a[n+1]+a[n+1]a[n+2]>a[n+2]a[n+3]∴rq^(n-1)+rq^n>rq^(n+1)1+q>q^2即:q^2-q...