设Sn为等差数列an的前n项和.求证Sn/n为等差数列
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设Sn为等差数列an的前n项和.求证Sn/n为等差数列
使用数学归纳法;设首项为a1,公差为d
当n=1时,S1=a1;当n=2时,S2/2=(a1+a1+d)/2=a1+d/2
当n=3时,S3/3=(a1+a1+d+a1+2d)/3=a1+d.
此时S3/3-S2/2=S2/2-S1=d/2
假设当n=k时,Sk/k-S(k-1)/(K-1)=d/2
当n=k+1时,
S(k+1)/(k+1)-Sk/k
=〔(k+1)a1+(d+2d+...+kd)〕/(K+1)-〔ka1+d+2d+...+(k-1)d〕/k
=a1+kd/2-〔a1+(k-1)d/2〕=d/2
因此各项差都是d/2,为等差数列.
当n=1时,S1=a1;当n=2时,S2/2=(a1+a1+d)/2=a1+d/2
当n=3时,S3/3=(a1+a1+d+a1+2d)/3=a1+d.
此时S3/3-S2/2=S2/2-S1=d/2
假设当n=k时,Sk/k-S(k-1)/(K-1)=d/2
当n=k+1时,
S(k+1)/(k+1)-Sk/k
=〔(k+1)a1+(d+2d+...+kd)〕/(K+1)-〔ka1+d+2d+...+(k-1)d〕/k
=a1+kd/2-〔a1+(k-1)d/2〕=d/2
因此各项差都是d/2,为等差数列.
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