数列an满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2.
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数列an满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2.3.求an的通项公式
第2问,设bn=a(2n-1)/a(2n),Sn=b1+b2+.....+bn,证明当n≥6时,|Sn-2|
第2问,设bn=a(2n-1)/a(2n),Sn=b1+b2+.....+bn,证明当n≥6时,|Sn-2|
![数列an满足a1=1,a2=2,a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)],n=1.2.](/uploads/image/z/6488383-31-3.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Ca2%3D2%2Ca%28n%2B2%29%3D%5B1%2Bcos%5E2%28n%CF%80%2F2%29%5Dan%2Bsin%5E2%28n%CF%80%2F2%29%5D%2Cn%3D1.2.)
a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)]
若n为偶数
a(n+2)=2an
若n为奇数
a(n+2)=an+1
∴a1,a3,a5……形成等差数列
a2,a4.26……形成等比数列
故而
当n为偶数时an=2^(n/2)
当n为奇数时an=(n+1)/2
2)
∵2n-1为奇数,2n为偶数
bn=n/(2^n)
用一下错位相减
Sn=2-(1/2)^n*(2+n)
|Sn-2|=(1/2)^n*(2+n)
设f(n)=|Sn-2|*n=(1/2)^n(2n+n^2)
f(n+1)=(1/2)^(n+1)(n^2+4n+3)
f(n+1)/f(n)=1/2*[1+(2n+3)/(n^2+2n)]
设t=2n+3≥15
f(n+1)/f(n)=1/2*[1+(2n+3)/(n^2+2n)]=1/2*[1+4/(t-2-3/t)]
∵4/(t-2-3/t)单调递减
∴4/(t-2-3/t)≤5/16(t=15)
f(n+1)/f(n)
若n为偶数
a(n+2)=2an
若n为奇数
a(n+2)=an+1
∴a1,a3,a5……形成等差数列
a2,a4.26……形成等比数列
故而
当n为偶数时an=2^(n/2)
当n为奇数时an=(n+1)/2
2)
∵2n-1为奇数,2n为偶数
bn=n/(2^n)
用一下错位相减
Sn=2-(1/2)^n*(2+n)
|Sn-2|=(1/2)^n*(2+n)
设f(n)=|Sn-2|*n=(1/2)^n(2n+n^2)
f(n+1)=(1/2)^(n+1)(n^2+4n+3)
f(n+1)/f(n)=1/2*[1+(2n+3)/(n^2+2n)]
设t=2n+3≥15
f(n+1)/f(n)=1/2*[1+(2n+3)/(n^2+2n)]=1/2*[1+4/(t-2-3/t)]
∵4/(t-2-3/t)单调递减
∴4/(t-2-3/t)≤5/16(t=15)
f(n+1)/f(n)
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