已知数列an的首项a1不等于0,公差d不等于0,的等差数列,求Sn=1./a1a2+1/a2a3+.+1/ana(n+1)
人气：180 ℃ 时间：2020-01-28 00:59:52
解答
因为1/anan+1=1/an*（an+d）=1/d[1/an-1/（an+d）]=1/d[1/an-1/an+1]
所以1/a1a2+1/a2a3+…+1/anan+1
=1/d[1/a1-1/a2+1/a3-1/a4.1/an-1/an+1]
=1/d（1/a1-1/an+1）
=1/d*（an+1-a1）/a1an+1
=n/a1(a1+nd)
祝您学习愉快
推荐
- 已知数列an为首项a1≠0,公差为d≠0的等差数列,求Sn=1/a1a2+1/a2a3+……+1/ana(n+1)
- （1）等差数列{1/an}满足a1=1,公差d=2,求a1a2+a2a3+...+ana(a+1)的和
- {an}为等差数列,an不等于0,d为公差,求证:1/(a1a2)+1/(a2a3)+...+1/(an-1*an)=(n-1)/(aian)
- 设{an}是等差数列,且首项a1>0,公差d>0求证:1/a1a2+1/a2a3+…+1/anan+1=n/a1(a1+nd)
- an是首项为3,公差,公差为2的等差数列,则lim(1/a1a2+1/a2a3+……+1/a(n-1)an)=
- city life isvery-------(difference)from the country life
- 在森林里遇到打雷下雨天躲在大树下安全吗?如果不安全,那该怎么办?
- People climb mountains at the Double Ninth Festival. climb mountains 划线提问
猜你喜欢
