已知数列a1=1,an*a(n+1)=(1/2)^n.1,求证数列{a2n}与{a(2n-1)}n?N*都是等比数列.2,求数列{an}的前2n项和T2n

Question

已知数列a1=1,an*a(n+1)=(1/2)^n.1,求证数列{a2n}与{a(2n-1)}n?N*都是等比数列.
2,求数列{an}的前2n项和T2n

Answer 1

an*a(n+1)=(1/2)^n.
a(n+1)*a(n+2)=(1/2)^(n+1)
2式除以1式得
a(n+2)/an=2
数列{a2n}与{a(2n-1)}n?N*都是等比数
T2n =S奇+S偶
=(1+2+4+ … )+(2+4+8+… ) 每个括号内都有n项
=3(1-2^n)/(1-2)
=3*2^n-3