已知数列的前n项和Sn=n^2+1/2n，求an并证明｛an｝是等差数列
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a1=S1=1^2+1/2*1=3/2
Sn=n^2+n/2
S(n-1)=(n-1)^2+(n-1)/2
=n^2-2n+1+n/2-1/2
=n^2-3n/2+1/2
an=Sn-S(n-1)
=n^2+n/2-(n^2-3n/2+1/2)
=n^2+n/2-n^2+3n/2-1/2
=2n-1/2
a(n-1)
=2(n-1)-1/2
=2n-2-1/2
=2n-5/2
an-a(n-1)
=2n-1/2-(2n-5/2)
=2n-1/2-2n+5/2
=2
所以an是以a1=3/2，公差为2的等差数列
Sn=n^2+n/2
S(n-1)=(n-1)^2+(n-1)/2
=n^2-2n+1+n/2-1/2
=n^2-3n/2+1/2
an=Sn-S(n-1)
=n^2+n/2-(n^2-3n/2+1/2)
=n^2+n/2-n^2+3n/2-1/2
=2n-1/2
a(n-1)
=2(n-1)-1/2
=2n-2-1/2
=2n-5/2
an-a(n-1)
=2n-1/2-(2n-5/2)
=2n-1/2-2n+5/2
=2
所以an是以a1=3/2，公差为2的等差数列
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