已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、设Tn=(1+a1)(1+a2)...(1+an),求Tn
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 16:46:29
![已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、设Tn=(1+a1)(1+a2)...(1+an),求Tn](/uploads/image/z/2778199-7-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D2%2Ca%EF%BC%88n%2B1%EF%BC%89%3Dan%5E2%2B2%2Aan.1%E3%80%81%E6%B1%82%E8%AF%81%EF%BC%9Blg%EF%BC%881%2Ban%EF%BC%89%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.2%E3%80%81%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D2%2Ca%EF%BC%88n%2B1%EF%BC%89%3Dan%5E2%2B2%2Aan.1%E3%80%81%E6%B1%82%E8%AF%81%EF%BC%9Blg%EF%BC%881%2Ban%EF%BC%89%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97.2%E3%80%81%E8%AE%BETn%3D%281%2Ba1%29%281%2Ba2%29...%281%2Ban%29%2C%E6%B1%82Tn)
已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、设Tn=(1+a1)(1+a2)...(1+an),求Tn
已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、
已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.
2、设Tn=(1+a1)(1+a2)...(1+an),求Tn
已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、已知数列an中,a1=2,a(n+1)=an^2+2*an.1、求证;lg(1+an)是等比数列.2、设Tn=(1+a1)(1+a2)...(1+an),求Tn
1.
∵a(n+1)=(an)^2+2an
∴a(n+1)+1=(an+1)^2.(1)
又∵a1=2>1
易之an>0
∴对(1)两边取常用对数,则:
lg[a(n+1)+1]=2lg(an+1)
又∵an+1≠1
∴lg[a(n+1)+1]/lg(an+1) = 2
即数列{lg(an+1)}是公比为2,首项为lg3的等比数列
2.
lg(an+1)=lg3 * (2^n -1)
lgTn=lg(1+a1)+lg(1+a2)+...+lg(1+an)
=lg3*[2-1+2^2-1+...+2^n-1]
=lg3*[2^(n+1)-n-2]
∴Tn=3^[2^(n+1)-n-2]
确认一下是不是log3(1+an),如果是,答案如下
a(n+1)=an^2+2*an
所以a(n+1)+1=an^2+2*an+1=(an+1)^2
即:an +1=[a(n-1)+1]^2=[a(n-2)+1]^(2^2)
=.........
=(a1+1)^[2^(n-1)]
...
全部展开
确认一下是不是log3(1+an),如果是,答案如下
a(n+1)=an^2+2*an
所以a(n+1)+1=an^2+2*an+1=(an+1)^2
即:an +1=[a(n-1)+1]^2=[a(n-2)+1]^(2^2)
=.........
=(a1+1)^[2^(n-1)]
=3^[2^(n-1)]
所以log3(1+an)/log3[1+a(n-1)]=2^(n-1)/2^(n-2)=2
所以log3(1+an)为等比数列
Tn=3*(3^2)*(3^4)*......*3^[2^(n-1)]
=3^[1+2+4+......+2^(n-1)]
=3^(2^n-1)
喔,楼上对的,我错了
收起