设Sn为等差数列前n项和,求证:{Sn/n}是等差数列

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设Sn为等差数列前n项和,求证:{Sn/n}是等差数列

设Sn为等差数列前n项和,求证:{Sn/n}是等差数列
设Sn为等差数列前n项和,求证:{Sn/n}是等差数列

设Sn为等差数列前n项和,求证:{Sn/n}是等差数列
Sn=n*a1+1/2*n*(n-1)*d
Sn/n=a1+1/2*(n-1)*d
初值为a1,公差为1/2*d

Sn=dn^2/2+(a1-d/2)n
所以:Sn/n=dn/2+a1-d/2
所以为等差数列

Sn/n=a1+(n-1)(d/2);
那么Sn/n是以a1为首项,d/2为等差的等差数列。

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