设数列{an}的前n项和为Sn,a1=1,an=sn/n+2(n-1),求证数列{an}是等差数列,并求其通项公式an

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设数列{an}的前n项和为Sn,a1=1,an=sn/n+2(n-1),求证数列{an}是等差数列,并求其通项公式an

设数列{an}的前n项和为Sn,a1=1,an=sn/n+2(n-1),求证数列{an}是等差数列,并求其通项公式an
设数列{an}的前n项和为Sn,a1=1,an=sn/n+2(n-1),求证数列{an}是等差数列,并求其通项公式an

设数列{an}的前n项和为Sn,a1=1,an=sn/n+2(n-1),求证数列{an}是等差数列,并求其通项公式an
an=sn/n+2(n-1)
Sn=nan-2n(n-1)
S(n-1)=(n-1)a(n-1)-2(n-1)(n-2)
Sn-S(n-1)=an=nan-2n(n-1)-[(n-1)a(n-1)-2(n-1)(n-2)]
=nan-2n(n-1)-(n-1)a(n-1)+2(n-1)(n-2)
=nan-2n(n-1)-na(n-1)+a(n-1)+2n(n-1)-4(n-1)
=n[an-a(n-1)]+a(n-1)-4(n-2)
n[an-a(n-1)]-[an-a(n-1)]=4(n-1)
[an-a(n-1)](n-1)=4(n-1)
an-a(n-1)=4
所以数列{an}是等差数列,公差d=4
An=a1+(n-1)d=1+(n-1)*4=4n-3

a[n]=S[n]/n+2(n-1)
na[n]=S[n]+2n(n-1)
(n-1)a[n]=S[n]-a[n]+2n(n-1)=S[n-1]+2n(n-1)
a[n]=S[n-1]/(n-1)+2n ------------(1)
同时因为a[n]=S[n]/n+2(n-1)
有a[n-1]=S[n-1]/(n-1)+2(n-2) ----...

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a[n]=S[n]/n+2(n-1)
na[n]=S[n]+2n(n-1)
(n-1)a[n]=S[n]-a[n]+2n(n-1)=S[n-1]+2n(n-1)
a[n]=S[n-1]/(n-1)+2n ------------(1)
同时因为a[n]=S[n]/n+2(n-1)
有a[n-1]=S[n-1]/(n-1)+2(n-2) ------(2)
(1)-(2),得
a[n]-a[n-1]=4
所以{an}是等差数列,且公差为4,这样
a[n]=a[1]+4*(n-1)=4n-3

收起

an=sn/n+2(n-1)得Sn=nan-2n(n-1),利用an=S(n)-S(n-1) (n>1)及a1=1,得到:(n-1)an-(n-1)a(n-1)-4(n-1)=0,即an-a(n-1)=4=常数,从而此数列为等差数列,且公差为4,得:an=4n-3。

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