Question

已知数列{an}，a1=1,anan+1=(1/2)^n(n∈N+)

(1)求证：数列{a2n}{a2n-1}（n∈n+）都是等比数列;
(2)求数列{an}的前2n项和T2n

Answer 1

用归纳法证明:
a(2n)=a(2n+1)=(1/2)^n,

N=1时,1/2=a(1)a(2)=a(2), (1/2)^2=a(2)a(3), a(3)=(1/2). 结论正确.
设n=k时,结论正确.也即,a(2k)=a(2k+1)=(1/2)^k.
当睁悉n=k+1时,(1/2)^(2k+1)=a(2k+1)a(2k+2), a(2k+2)=(1/2)^(k+1).
(1/2)^(2k+2)=a(2k+2)a(2k+3), a(2k+3)=(1/2)^(k+1)=a(2k+2).结论正确.
因此,总有
a(2n)=a(2n+1)=(1/2)^n, a(1)=1.
{a(2n)},{a(2n-1)}都是公比为(1/2)的等比数列.

T(2n)=a(1)+a(3)+…+a(2n-1) + a(2)+a(4)+…+a(2n)
=1+(1/2)+…+(1/2)^(n-1) + (1/2)+(1/悉埋乎2)^2+…+(1/2)^n
=3/液册2[1+1/2+…+1/2^(n-1)]
=3[1-1/2^n]