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

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

cccc分 1年前 已收到2个回答 举报

迷恋灵魂 幼苗

共回答了9个问题采纳率:88.9% 举报

(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

1年前

3

粉色梦境 幼苗

共回答了760个问题 举报

a1=1

a(n+1)={(1/2)an+n(n是奇数)

{an-2n(n是偶数)

................................................................................

(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:

b(n+1)=(a(2n+2)-4n-4)+4n+2=a(2n+2)-2

a(2n+2)=a[(2n+1)+1]=(1/2)a(2n+1)+(2n+1)

b(n+1)=(1/2)a(2n+1)+2n-1=(1/2)[a(2n+1)+4n-2]=(1/2)bn

b(n+1)/bn=(1/2)=q

b1=a3+2

a3=a2-4

b1=a2-2

a2=(1/2)+1=3/2

b1=-(1/2)

bn= -((1/2)^n

(2)

a(2n+1)=a2n-4n

a2n=a[(2n-1)+1]=(1/2)a(2n-1)+(2n-1),代入上式得:

a(2n+1)=[(1/2)a(2n-1)+(2n-1)]-4n=(1/2)a(2n-1)-(2n+1)

令bn=a(2n-1)(以下暴难)

b(n+1)=(1/2)bn-(2n+1)

b(n+1)+[4(n+1)-10]=(1/2){bn+(4n-10)}

再令,cn=bn+(4n-10)

c(n+1)=(1/2)cn

c(n+1)/cn=1/2=q

c1=b1-6=a1-6=-5

所以等比数列:cn= -5*(1/2)^(n-1)

原数列的前100个奇数项是{cn}的前50项

c1+c2+....+c50=10[(1/2)^(50)-1]

以下是用图片共两张:

(1)

(2)

1年前

2
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