数列an中a1=1 ,an*a(n+1)=4的n次幂,s2n=?

数列an中a1=1 ,an*a(n+1)=4的n次幂,s2n=?
数列an中a1=1 ,an*a(n+1)=4的n次幂,s2n=?

数列an中a1=1 ,an*a(n+1)=4的n次幂,s2n=?
a(n+1)=4^n/an=4^n/(4^(n-1)/a(n-1))=4a(n-1)
a1=1
a2=4/a1=4
a3=4a1=4
a4=4a2=4^2
……
a(2n-1)=4^n-1
a(2n)=4^n
s2n=1+4+……4^(n-1)+4+4^2+……4^n=5(4^n-1)/3
方法是这样,没拿笔算你自己看看对不对