tfsf.net
当前位置:首页>>关于在数列{an}中,已知a1=-1,且a(n+1)=2an+3n-4,求...的资料>>

在数列{an}中,已知a1=-1,且a(n+1)=2an+3n-4,求...

这道题要利用 数列中的迭代方法……步骤很麻烦.方便的话把你的邮箱给我,我再word上做完,发给你!在word上可以利用数学编辑器,在百度里不能用!

由1、2可解出,an=2^(n-1)-3*n+1,sn=2^n-(3/2)*n*n-(1/2)*n-1易知,当n>=5时,an>0a1=-1,a2=-3,a3=-4,a4=-3Sn=|a1|+|a2|+|a3|+.+|an|=sn-2*s4=22+sn

设 bn= a(n+1)-an+3由a(n+1)=2an+3n-4得 bn= an+3n -1又由由a(n+1)=2an+3n-4得 a(n+1) + 3(n+1 )-1 =2an+6n-2 = 2( an +3n -1 )即 b(n+1)=2bn{bn} 是 公比为2的等比数列, b1= a1+2 =1bn = b1* 2^(n-1) = 2^(n-1)an = 2^(n-1) -3n+1

1,a(n+2)=2a(n+1)+3n-1,与a(n+1)=2an+3n-4作差,得a(n+2)=3a(n+1)-2an+3,整理得a(n+2)-a(n+1)+3=2*{a(n+1)-an+3},又因为a1=-1,所以数列{a(n+1)-an+3}是等比数

(1)a(n+1)=2an+3n-4a(n+1)+3n-4=2(an+3n-4)所以数列{an+3n-4}是首项为a1+3-4=-2,公比为q=2的等比数列an+3n-4=-2^nan=-2^n-3n+4an+1-an+3=-2^n所以数列{an+1-an+3}是首项是否2,公比为2的等比数列(2){an}的通项公式:an=-2^n-3n+4(3)=……

a2=2a1+3+4=9a(n+1)=2an+3n+4an=2a(n-1)+3(n-1)+4a(n+1)-an=2[an-a(n-1)+3a(n+1)-an+3=2[an-a(n-1)+3]故{a(n+1)-an+3}是以a2-a1+3=9-1+3=11为首项,公比是2的等比数列.即有a(n+1)-an+3=11*2^(n-1)a(n+1)-an=11*2^(n-1

1)a1=入,a2=2入+2,a3=4入+9,很明显an不可能是等比数列2)bn=an+3n-1;b(n+1)=a(n+1)+3(n+1)-1=2an+3n-4+3n+2=2(an+3n-1)=2bn,所以bn为等比数列,公比为2 3)算出bn之后,再反过来求an

(1)∵bn=an+3n-1,∴bn+1=(an+1)+3(n+1)-1=(2an+3n-4)+3n+2=2an+6n-2=2(an+3n-1)=2bn∴{bn}是以2为公比,b1=a1+3-1=3为首项的等比数列.∴bn=3*2^(n-1)(2)∵bn=an+3n-1=3*2^(n-1)∴an=3*2^(n-1)-3n+1,利用分组求和,就是{3*2^(n-1)}、{-3n}、{1}三个数列的和∴Sn=3(1-2^n)/(1-2)-3n(n+1)/2+n=32^-3n(n+1)/2+n-3

1,a(n+2)=2a(n+1)+3n-1,与a(n+1)=2an+3n-4作差,得a(n+2)=3a(n+1)-2an+3,整理得a(n+2)-a(n+1)+3=2*{a(n+1)-an+3},又因为a1=-1,所以数列{a(n+1)-an+3}是等比数列 2.a1=-1带入a(n+1)=2an+3n-4(n∈N*),得a2=-3,a3=-4设cn=a(n+1)-an+

相关文档
网站首页 | 网站地图
All rights reserved Powered by www.tfsf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com