(1)已知数列an的前n项和为sn满足sn=an²+bn,求证an是等差数列
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(1)已知数列an的前n项和为sn满足sn=an²+bn,求证an是等差数列
(2)已知等差数列an的前n项和为sn,求证数列sn/n也成等差数列
(2)已知等差数列an的前n项和为sn,求证数列sn/n也成等差数列
Sn=an²+bn
则:
当n=1时,a1=S1=a+b
当n≥2时,an=Sn-S(n-1)=[an²+bn]-[a(n-1)²+b(n-1)]=2an-(a-b),其中n≥2
当n=1时,也满足上式.
得:an=2an-(a-b)
当n≥2时,an-a(n-1)=2a=常数.
则数列{an}是等差数列.
Sn/n=an+n
则:[S(n+1)/(n+1)]-[Sn/n]=a=常数
数列{Sn/n}也是等差数列.
则:
当n=1时,a1=S1=a+b
当n≥2时,an=Sn-S(n-1)=[an²+bn]-[a(n-1)²+b(n-1)]=2an-(a-b),其中n≥2
当n=1时,也满足上式.
得:an=2an-(a-b)
当n≥2时,an-a(n-1)=2a=常数.
则数列{an}是等差数列.
Sn/n=an+n
则:[S(n+1)/(n+1)]-[Sn/n]=a=常数
数列{Sn/n}也是等差数列.
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