已知数列满足：A1=1.AN+1=1/2AN+N,N奇数,AN-2N.N偶数

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/15 07:05:05

已知数列满足：A1=1.AN+1=1/2AN+N,N奇数,AN-2N.N偶数
射BN=A2N+4N-2,证明数列BN是等比数列
数列AN前100项,所以奇数项的和S

（1）
bn=a(2n+1)+4n-2
b(n+1)=a(2n+3)+4(n+1)-2=a(2n+2+1)+4n+2
=a(2n+2)-2(2n+2)+4n+2
=a(2n+1+1)-2(2n+2)+4n+2
1/2a(2n+1)+2n+1-2(2n+2)+4n+2
=1/2a(2n+1)+2n-1
=1/2[a(2n+1)+4n-2]
∴b(n+1)/bn=1/2
∴数列{bn}是公比为1/2的等比数列
b1=a3+2=a2-4+2=1/2a1+1-2=-1/2
bn=-(1/2)ⁿ
（2）
∵bn=a(2n+1)+4n-2
∴a(2n+1)=bn-4n+2=-1/2ⁿ-4n+2
S=a1+a3+a5+.+a99
=1+(-1/2-1/4-1/8-...-1/2^49)-4(1+2+3+...+49)+2×49
=1-(1-1/2^49)-2×49×50+98
= 1/2^49-4802

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