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.

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)