已知数列{an}中,a1=1且an*an+=2^(1⼀4)^n求通项式

an*a(n+1)=2^(1/4)^n (1)
put n=1
a2= 2^(1/4)
a(n-1).an = 2^(1/4)^(n-1) (2)

(1)/(2)
a(n+1)/a(n-1) = 2^(-(3/4). (1/4)^(n-1))
an/a(n-2) = 2^(-(3/4). (1/4)^(n-2))
if n is odd number
an/a1= 2^( (-3/4)[ (1/4)^1+ (1/4)^3+..+(1/4)^(n-2) ]
an = 2^(-(1/5)[ 1- (1/4)^(n-1) ])
=2^( (4/5)(1/4)^(n)- 1/5 )
if n is even number
an/a(n-2) = 2^(-(3/4). (1/4)^(n-2))
an/a2 =2^( (-3/4)[ (1/4)^2+(1/4)^4+..+(1/4)^(n-2)] )
= 2^( -(1/20)[ 1- (1/4)^(n-2) ] )
an = 2^( -(1/20)[ 1- (1/4)^(n-2) ] ) .2^(1/4)
= 2^((4/5)(1/4)^n+ 1/5 )

an =2^( (4/5).(1/4)^n-1/5) ) ；n 是奇数
= 2^((4/5).(1/4)^n+1/5 ) ; n 是偶数

Sn=(an+1)^2/4=(an^2+2an+1)/4
Sn-1=[a(n-1)+1]^2=[(a(n-1)^2+2a(n-1)+1]/4

Sn-Sn-1=an=[an^2+2an-a(n-1)^2-2a(n-1)]/4
4an=an^2+2an-a(n-1)^2-2a(n-1)
an^2-2an=a(n-1)^2+2a(n-1)
(an-1)^2=[a(n-1)+1]^2
又a1=1，an>0
an-1=a(n-1)+1
an-a(n-1)=2

数列为等差数列，首项为1，公差为2。
an=1+(n-1)*2=2n+1

等式两边取对数即可