已知数列{an}满足,a1=2,a(n+1)=3根号an,求通项an

a1=2>0
假设当n=k(k∈N+)时,ak>0,则a(k+1)=3√ak >0
k为任意正整数,因此对于任意正整数n,an恒>0,数列各项均为正.
a(n+1)=3√an
log3[a(n+1)]=log3(3√an)=(1/2)log3(an) +1
log3[a(n+1)] -2=(1/2)log3(an) -1=(1/2)[log3(an) -2]
{log3[a(n+1)] -2}/[log3(an) -2]=1/2,为定值.
log3(a1) -2=log3(2) -2,数列{log3(an) -2}是以log3(2) -2为首项,1/2为公比的等比数列.
log3(an) -2=[log3(2) -2]×(1/2)^(n-1)=log3[(2/9)^(1/2)^(n-1)]
log3(an/9)=log3[(2/9)^(1/2)^(n-1)]
an=9× (2/9)^[(1/2)^(n-1)]
数列{an}的通项公式为an=9× (2/9)^[(1/2)^(n-1)]
√Sn=an +1/4
Sn=(an +1/4)^2
n=1时,a1=S1=(a1+ 1/4)^2
整理,得(a1 -1/4)^2=0
a1=1/4
n≥2时,an=Sn-S(n-1)=(an +1/4)^2 -[a(n-1)+1/4]^2
2an^2- 2a(n-1)^2 -an -a(n-1)=0
2[an+a(n-1)][an-a(n-1)]-[an+a(n-1)]=0
[an+a(n-1)][an -a(n-1) -1/2]=0
an>0 an+a(n-1)>0,因此只有an-a(n-1)=1/2,为定值.
数列{an}是以1/4为首项,1/2为公差的等差数列
an=(1/4)+(1/2)(n-1)=n/2 -1/4
数列{an}的通项公式为an=n/2 -1/4