Question

是否存在等差数列｛An｝,使对任意n,都有An*Sn=2n^2(n+1)

Answer 1

存在

假设存在这样的等差数列an

设首项为a1,公差为d

an=a1+(n-1)d

Sn=na1+n(n-1)d/2

anSn=[a1+(n-1)d][na1+n(n-1)d/2]=2n^2(n+1)

根据三系系数可知d=2

∴anSn=[a1+2(n-1)][na1+n(n-1)]=n[2n+a1-2][n+a1-1]=n[2n^2+n(3a1-4)+(a1-1)(a1-2)]=n[2n^2+2n]

∴3a1-4=2

(a1-1)(a1-2)]=0

∴a1=2

即an=2+2(n-1)=2n

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