已知数列{an}满足:a1=1,an+1=1/2an+n,n 为奇数,an-2n,n 为偶数.设bn=a2n+1+4n-2,n€N

bn=a(2n+1)+4n-2
b(n+1)=a(2n+3)+4n+2
=a(2n+2)-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}是等比数列,公比为1/2
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=-1/2^n-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

已知数列{an}满足a1=1,an+1=｛1/2an+n,n为奇数，an-2n，n为偶数｝ 记bn=a2n Tn=a1+a2+a3+……a（2n+1）~

a2=1/2*a1+1=1.5
a(2n)=1/2*a(2n-1)+2n-1
=1/2*[a(2n-2)-2(2n-2)]+2n-1
=1/2*a(2n-2)+1
设cn=bn-2=a(2n)-2
c1=a2-2=1.5-2=-0.5
c(n-1)=a(2n-2)-2即a(2n-2)=c(n-1)+2
cn=a(2n)-2=1/2*a(2n-2)+1-2=1/2*[c(n-1)+2]+1-2=1/2*c(n-1)
所以cn是首项为c1=-0.5，公比为q=1/2的等比数列
其通项公式为
cn=c1*q^(n-1)=-0.5*(1/2)^(n-1)=-1/(2^n)
其前n项和为
Sn=c1*(1-q^n)/(1-q)=1/(2^n)-1
所以bn=cn+2=2-1/(2^n)

a(2n+1)=a(2n)-2(2n)=a(2n)-4n
a(2n)+a(2n+1)=2a(2n)-4n=2(cn+2)-4n=2cn-4(n-1)
所以
Tn=a1+a2+a3+……a（2n+1）
=a1+(a2+a3)+...+[a(2n)+a(2n+1)]
=1+2c1+...+[2cn-4(n-1)]
=1+2(c1+c2+...+cn)-4[1+2+...+(n-1)]
=1+2Sn-4n(n-1)/2
=1+[2/(2^n)-2]-2n(n-1)
=1/[2^(n-1)]-1-2n(n-1)

(I)
a(n+1)=an +1 ,n is odd
=2an ,n is even
bn=a(2n) -1

a1=1
a2 =a1+1 =2

if n is odd,
a(n+1) = an +1
= 2a(n-1) +1
a(n+1) +1 = 2[ (a(n-1) +1 ]
a(n+1) +1 = 2^[( n-1)/2 ]. (a2 +1 )
= 3.2^[( n-1)/2 ]
a(n+1) = -1+3.2^[( n-1)/2 ]

n is odd => n= 2m-1
a(2m) =-1+3.2^(m-1)

bn = a(2n) -1
=-2+3.2^(n-1)
b2 = -2+3.2 = 4
b3= -2+3.4= 10

b(n+1) =-2+3.2^n
=2(-2+3.2^(n-1)) +2
=2bn +2
(II)
(1)

bn+2 =3.2^(n-1)
{bn+2}是等比数列, q=3

(2)
if n is even,
a(n+1)=2an
= 2(a(n-1)+1)
a(n+1)+2 =2(a(n-1)+2)
= 2^(n/2). (a1+2)
=3. 2^(n/2)
a(n+1) =-2+3.2^(n/2)

n is even , n=2k
a(2k+1) =-2+3.2^k

a(2k),a(2k+1),9+a(2k+2)成等比数列
a(2k).[9+a(2k+2)]= [a(2k+1)]^2
[-1+3.2^(k-1) ].(8+3.2^k ) =(-2+3.2^k)^2
-8-3.2^k+12.2^k +9.2^(2k-1) = 4 -12.2^k + 9.2^(2k)
(9/2).2^(2k) -21.2^k +12=0
9.2^(2k) -42.2^k +24=0
(2^k -4)(9.2^k-6)=0
2^k=4
k=2

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