已知数列{an}满足an+1+an=4n-3 当a1=2时,求Sn
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已知数列{an}满足an+1+an=4n-3 当a1=2时,求Sn
![已知数列{an}满足an+1+an=4n-3 当a1=2时,求Sn](/uploads/image/z/5783852-20-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3an%2B1%2Ban%3D4n-3+%E5%BD%93a1%3D2%E6%97%B6%2C%E6%B1%82Sn)
a(n+1)+an=4n-3,an+a(n-1)=4*(n-1)-3,
故a(n+1)-a(n-1)=4,(n≥2)
a1=2,a2=-1
当n为奇数时,
an=2+(n-1)/2*4=2n,a(n-1)=-1+(n-1)/2*4=2n-5,
故Sn=(2+2n)*(n+1)/2/2+(-1+2n-5)*(n-1)/2/2
=n^2-n+2
当n为偶数时,
an=-1+2n,a(n-1)=2+2n,
故Sn=(2+2+2n)*n/2/2+(-1+2n-1)*n/2/2=n^2+n/2
故a(n+1)-a(n-1)=4,(n≥2)
a1=2,a2=-1
当n为奇数时,
an=2+(n-1)/2*4=2n,a(n-1)=-1+(n-1)/2*4=2n-5,
故Sn=(2+2n)*(n+1)/2/2+(-1+2n-5)*(n-1)/2/2
=n^2-n+2
当n为偶数时,
an=-1+2n,a(n-1)=2+2n,
故Sn=(2+2+2n)*n/2/2+(-1+2n-1)*n/2/2=n^2+n/2
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