等差数列{an}前n项和为sn,公差为d=3),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)
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![等差数列{an}前n项和为sn,公差为d=3),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)](/uploads/image/z/2137311-63-1.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E5%85%AC%E5%B7%AE%E4%B8%BAd%3D3%29%2C%E4%BD%BF%E5%BE%97am%3Dsm%2C%E5%88%99%E5%BD%93n%3Em%28n%E2%88%88N%2A%29%E6%97%B6sn%E5%92%8Can%28%E5%A1%AB%2C%3D%29)
等差数列{an}前n项和为sn,公差为d=3),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)
等差数列{an}前n项和为sn,公差为d=3),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)
等差数列{an}前n项和为sn,公差为d=3),使得am=sm,则当n>m(n∈N*)时sn和an(填,=)
用图像最好做!应该是 <
用代数方法是这样的:an=a1+(n-1)d
Sn=a1+[n(n-1)d]/2
作差得:Sn-an=(n/2-1)(n-1)d
由题知,n>m>=3 ,d<0 所以Sn-an<0
即Sn
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