搜搜做题作业网等差数列 2到填空题
来源：学生作业帮 编辑：搜搜做题作业网作业帮 分类：数学作业 时间：2024/07/09 05:41:23
等差数列 2到填空题
{an}与{bn}是等差数列
∴Sn=[n(a1+an)]/2
Tn=[n(b1+bn)]/2
∴Sn/Tn=(a1+an)/(b1+bn)
∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)
∴(a1+an)/(b1+bn)=2n:(3n+1)
假设(n+1)/2 =k {(n+1)/2为项数}
则n=2k-1
则ak/bk = 2(2k-1)/[3(2k-1)+1]
=(2k-1)/(3k-1)
即an/bn =(2n-1)/(3n-1)
sn=a1+a2+...an=1og2(2/3)+1og2(3/4)+.log2(n+1/n+2)=log2(2/3x3/4x.n+1/n+2)=log2(2/n+2)
∴Sn=[n(a1+an)]/2
Tn=[n(b1+bn)]/2
∴Sn/Tn=(a1+an)/(b1+bn)
∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)
∴(a1+an)/(b1+bn)=2n:(3n+1)
假设(n+1)/2 =k {(n+1)/2为项数}
则n=2k-1
则ak/bk = 2(2k-1)/[3(2k-1)+1]
=(2k-1)/(3k-1)
即an/bn =(2n-1)/(3n-1)
sn=a1+a2+...an=1og2(2/3)+1og2(3/4)+.log2(n+1/n+2)=log2(2/3x3/4x.n+1/n+2)=log2(2/n+2)