数列{an}满足a1=1,a2=2,an+2=[1+cos2(nπ/2)]an+sin2(nπ/2)则该数列的前20项 5
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a(n+2)=(1+[cos(nπ/2)]^2)an+[sin(nπ/2)]^2
when n is even
a(n+2) = (1+[cos(nπ/2)]^2)an+[sin(nπ/2)]^2
= 2an
an/a(n-2) =2
an/a2 = 2^[(n-2)/2]
an = 2^(n/2)
ie
a2+a4+a6+...+a20 = 2^1+2^2+...+2^10
= 2(2^10-1)
when n is odd
a(n+2) = (1+[cos(nπ/2)]^2)an+[sin(nπ/2)]^2
= an + 1
an - a(n-2) =1
an -a1 = (n+1)/2 -1
an =(n+1)/2
ie
a1+a3+a5+...+a19 = 1 + 2+...+10
= 55
a1+a2+a3+...+a20 = 2(2^10-1) +55
= 53 + 2^11
when n is even
a(n+2) = (1+[cos(nπ/2)]^2)an+[sin(nπ/2)]^2
= 2an
an/a(n-2) =2
an/a2 = 2^[(n-2)/2]
an = 2^(n/2)
ie
a2+a4+a6+...+a20 = 2^1+2^2+...+2^10
= 2(2^10-1)
when n is odd
a(n+2) = (1+[cos(nπ/2)]^2)an+[sin(nπ/2)]^2
= an + 1
an - a(n-2) =1
an -a1 = (n+1)/2 -1
an =(n+1)/2
ie
a1+a3+a5+...+a19 = 1 + 2+...+10
= 55
a1+a2+a3+...+a20 = 2(2^10-1) +55
= 53 + 2^11
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当n为奇数时,cos(nπ/2)=0,sin(nπ/2)=±1
所以a(n+2)=an+1
当n为偶数时,cos(nπ/2)=±1,sin(nπ/2)=0
所以a(n+2)=2an
因此a3=2,a5=3,a7=4,...,a19=10
a4=4,a6=8,a8=16,...a20=2^10
所以S20=(1+2+3+...+10)+(2+4+8+...+2^10)
=55+2^11-2=2101
所以a(n+2)=an+1
当n为偶数时,cos(nπ/2)=±1,sin(nπ/2)=0
所以a(n+2)=2an
因此a3=2,a5=3,a7=4,...,a19=10
a4=4,a6=8,a8=16,...a20=2^10
所以S20=(1+2+3+...+10)+(2+4+8+...+2^10)
=55+2^11-2=2101
追问
再问一题。。。已知数列{an}满足a1+a2+...+an=n-an,其中n正整数
令bn=(2-n)(an-1)求数列bn的最大值
追答
Sn=n-an
S(n-1)=n-1-a(n-1)
两式相减,得an=1-an+a(n-1)
即an-1=1/2(a(n-1)-1)
所以{an-1}是等比数列,公比为1/2
而a1=1-a1,即a1=1/2,,a1-1=-1/2
所以an-1=-1/2^n
bn=(n-2)/2^n
当n=1时bn最小,最小值为b1=-1/2
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