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

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（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
======以下答案可供参考======
供参考答案1:
a1=1a(n+1)={(1/2)an+n                   (n是奇数）
            ｛an-2n                       (n是偶数）
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(1)bn=a(2n+1)+4n-2
b(n+1)=a(2n+3)+4n+2.............①
a(2n+3)=a[(2n+2)+1]=a(2n+2)-2(2n+2)=a(2n+2)-4n-4
把a(2n+3)的右边代入上式得