等差数列an=2n+3,求和:(1/a1a2)+(1/a2a3)+.+(1/anan+1)
人气:397 ℃ 时间:2020-02-03 08:30:08
解答
原式=1/(5×7)+1/(7×9)+1/(9×11)+.+1/[(2n+3)(2n+5)]
=1/2[(1/5-1/7)+(1/7-1/9)+(1/9-1/11)+.+1/(2n+3)-1/(2n+5)]
=1/2[1/5-1/(2n+5)]
=n/(10n+25)
推荐
- 已知数列{an},若1/a1a2+1/a2a3+…+1/anan-1=n/anan+1,求证{an}为等差数列.
- 已知:数列{an}的前n项和Sn=n2+2n(n∈N*) (1)求:通项 an (2)求和:1/a1a2+1/a2a3+1/a3a4+…+1/anan+1.
- 数列a1=1,an=an+1(1+2an)求证数列an分之一等差数列,若a1a2+a2a3+..+anan+1大于33分之16,求N的取值范
- 等差数列{1\an}满足a1=1,公差d=2,求a1a2+a2a3+……+anan+1的和
- 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)
- 玻璃制的镜子,最早出现在哪里,我国最早的玻璃镜子是
- 我们应该怎样珍惜生命、保护生物的生存环境?
- Well-established principles-both in this Circuit and else-where-furnish the analytical scaffolding for determining wheth
猜你喜欢