a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)]
若n為偶數(shù)
a(n+2)=2an
若n為奇數(shù)
a(n+2)=an+1
∴a1,a3,a5……形成等差數(shù)列
a2,a4.26……形成等比數(shù)列
故而
當(dāng)n為偶數(shù)時(shí)an=2^(n/2)
當(dāng)n為奇數(shù)時(shí)an=(n+1)/2
2)
∵2n-1為奇數(shù),2n為偶數(shù)
bn=n/(2^n)
用一下錯(cuò)位相減
Sn=2-(1/2)^n*(2+n)
|Sn-2|=(1/2)^n*(2+n)
設(shè)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)]
設(shè)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)單調(diào)遞減
∴4/(t-2-3/t)≤5/16(t=15)
f(n+1)/f(n)<1
f(n)單調(diào)遞減
f(n)max=f(6)=3/4<1
得證|Sn-2|<1/n