Question

等比数列｛an｝的各项均为正数，且2a1+3a2=1,a3²=9a2a6,求通项公式

Answer 1

(1)a3^2=9a2a6
(a2p)^2=9a2(a2p^4)
a2^2p^2=9a2^2p^4
∵此数列各项均为正数∴a2^2<>0,p>0
两边同时除以a2^2p^2，得9p^2=1,p=1/3
2a1+3a2=1
2a1+3*[(1/3)*a1]=1
2a1+a1=1
3a1=1
a1=1/3
an=a1p^(n-1)=1/3*(1/3)^(n-1)=1/3^n
(2)bn=log3a1+log3a2+...+log3an
=log3(a1*a2*...*an)
=log3[(1/3)*(1/3^2)*...*(1/3^n)]
=log3[(1/3)^(1+2+...+n)]
=(1+2+...+n)*log3(1/3)
=-n(n+1)/2
1/bn=-2/n(n+1)
=(-2)*[1/n(n+1)]
=(-2)*[1/n-1/(n+1)]
1/b1+1/b2+...+1/bn
=(-2)*(1-1/2)+(-2)*(1/2-1/3)+...+(-2)*[1/n-1/(n+1)]
=(-2)*[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=(-2)*[1-1/(n+1)]
=-2n/(n+1)

Answer 2

(1)a3^2=9a2a6
(a2p)^2=9a2(a2p^4)
a2^2p^2=9a2^2p^4
∵此数列各项均为正数∴a2^2<>0,p>0
两边同时除以a2^2p^2，得9p^2=1,p=1/3
2a1+3a2=1
2a1+3*[(1/3)*a1]=1
2a1+a1=1
3a1=1
a1=1/3
an=a1p^(n-1)=1/3*(1/3)^(n-1)=1/3^n

Answer 3

a3²=9a2a6
9q²=1
q=1/3
2a1+3a2=1
2a1+3a1*1/3=1
a1=1/3
an=(1/3)^n