等差数列{an}前n项和为sn,公差为d1),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)
1个回答

由am=Sm=m(a1+am)/2,得

2am=ma1+mam,

(2-m)am=ma1,

1)m=2时a1=0,an=d(n-1),Sn=dn(n-1)/2,

d2,

∴an>Sn.

2)m>2时am=ma1/(2-m),

an=am+(n-m)d=ma1/(2-m)+(n-m)d,

Sn=n(a1+an)/2=n[2a1/(2-m)+(n-m)d]/2,

∴2(Sn-an)=(2n-2m)a1/(2-m)+d(n-m)(n-2),

由am=ma1/(2-m)=a1+(m-1)d,得

ma1=(2-m)a1+(m-1)(2-m)d,

(2m-2)a1=-(m-1)(m-2)d,

m>2,d0,

∴an>Sn.

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