数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 20:39:22
![数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。](/uploads/image/z/3946026-66-6.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2Ca2%3D2%2Ca%28n%2B2%29%3D%5Ban%2Ba%28n%2B1%29%5D%2F2%2Cn%E5%B1%9E%E4%BA%8EN%2A.%E6%B1%82%281%29%E4%BB%A4bn%3Da%28n%2B1%29-an%EF%BC%8C%E8%AF%81%E6%98%8E%7Bbn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E3%80%82%282%29%E6%B1%82%7Ban%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E3%80%82%E6%B1%82%281%29%E4%BB%A4bn%3Da%28n%2B1%29-an%EF%BC%8C%E8%AF%81%E6%98%8E%7Bbn%7D%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E3%80%82%282%29%E6%B1%82%7Ban%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E3%80%82)
数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。
数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.
求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。
数列{an}满足a1=1,a2=2,a(n+2)=[an+a(n+1)]/2,n属于N*.求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。求(1)令bn=a(n+1)-an,证明{bn}是等比数列。(2)求{an}通项公式。
等式两边同时减去a(n+1){加(1/2)a(n+1)也行},然后后面的提出-1/2,然后就有很明显的规律了,一个双项的通项就出来了,是关于a(n+2)和a(n+1)的,然后就简单了,你先做做看,不懂再问.
2a(n+2)=an+a(n+1)同时减去2a(n+1)
∴2[a(n+2)-a(n+1)]=-[a(n+1)-an]
a(n+1)-an是以a2-a1=1为首项,以-1/2等比的等比数列
s(n-1)为数列和
an=1+s(n-1)
an=1+(2/3)[1-(-1/2)^(n-1)]=5/3+(1/3)×(-1/2)^(n-2)
就二楼的答案
If you would bother to bring God what to do ,Ray Ban sunglasses, because god is a girl she's only a girl do you believe it can you receive it