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
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
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]