Question

设各项均为正数的数列{an}的前n项和为Sn,已知2an=a1+a3 数列{根号Sn}是公差为d的等差数列

Answer 1

这是今年江苏卷上的题目…………

(1)

设根号Sn=d*n+H

Sn=d^2*n^2+2*d*H*n+H^2

a1=S1=d^2+2*d*H+H^2

a2=S2-S1=3*d^2+2*d*H

a3=S3-S2=5*d^2+2*d*H

a1+a3=2*a2

则H=0

Sn=d^2*n^2

n=1时

a1=S1=n^2

n>=2时

an=Sn-S(n-1)=(2n-1)*d^2

代入n=1成立

故an=(2n-1)*d^2

(2)

证明：

Sm=m^2*d^2

Sn=n^2*d^2

Sk=k^2*d^2

m+n=3k

Sm+Sn>c*Sk

Sm+Sn=d^2*(m^2+n^2)>c*k^2*d^2=c*(m+n)^2*d^2/9

由an各项为正数,an=(2n-1)*d^2,d不为0

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