已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Snnan+1的n+1是下角标哈.求证数列{Sn/n}为等比数列求an的通项公式及前N项和Sn
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![已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Snnan+1的n+1是下角标哈.求证数列{Sn/n}为等比数列求an的通项公式及前N项和Sn](/uploads/image/z/362663-71-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8DN%E9%A1%B9%E5%92%8C%E4%B8%BASn%2Ca1%3D1.nan%2B1%3D%28n%2B2%29Snnan%2B1%E7%9A%84n%2B1%E6%98%AF%E4%B8%8B%E8%A7%92%E6%A0%87%E5%93%88.%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%7BSn%2Fn%7D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E6%B1%82an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%8F%8A%E5%89%8DN%E9%A1%B9%E5%92%8CSn)
已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Snnan+1的n+1是下角标哈.求证数列{Sn/n}为等比数列求an的通项公式及前N项和Sn
已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Sn
nan+1的n+1是下角标哈.
求证数列{Sn/n}为等比数列
求an的通项公式及前N项和Sn
已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Snnan+1的n+1是下角标哈.求证数列{Sn/n}为等比数列求an的通项公式及前N项和Sn
nan+1=(n+2)Sn,则
(n-1)an=(n+1)S(n-1)
相减,
nan+1/(n+2)-(n-1)an/(n+1)=an
即
nan+1/(n+2)=2nan/(n+1)
则,
an+1/an=2(n+2)/(n+1)
则,
(Sn/n)/(Sn/(n-1))=(an+1/(n+2))/(an/(n+1))
=(n+1)/(n+2)*2(n+2)/(n+1)
=2
则{Sn/n}为等比数列
Sn/n=(1-2^n)/(1-2)=2^n-1
则,Sn=n(2^n-1)
则S(n-1)=(n-1)(2^(n-1)-1)
相减,
an=n(2^n-1)-(n-1)(2^(n-1)-1)
得
an=(n+1)*2^(n-1)+1
将An+1=Sn+1-Sn代入nAn+1=(n+2)Sn可得
n(Sn+1-Sn)=(n+2)Sn
即nSn+1=(2n+2)Sn
即Sn+1/n+1=2*Sn/n
则Sn/n是以S1/1=A1/1=1为首项,2为公比的等比数列
则Sn/n=2^(n-1),Sn=n*2^(n-1);
当n>=2时,An=Sn-Sn-1=n*2^(n-1)-(n-1)...
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将An+1=Sn+1-Sn代入nAn+1=(n+2)Sn可得
n(Sn+1-Sn)=(n+2)Sn
即nSn+1=(2n+2)Sn
即Sn+1/n+1=2*Sn/n
则Sn/n是以S1/1=A1/1=1为首项,2为公比的等比数列
则Sn/n=2^(n-1),Sn=n*2^(n-1);
当n>=2时,An=Sn-Sn-1=n*2^(n-1)-(n-1)*2^(n-2)=(n+1)*2^(n-2)
A1=1也满足上式,则An=(n+1)*2^(n-2),(n=1,2,3,。。。)
上面已求出Sn=n*2^(n-1)
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2楼答案正确 一楼错了
na(n+1)=(n+2)Sn
n(Sn+1-Sn)=(n+2)Sn
nSn+1=2(n+1)Sn
Sn+1/(n+1)=2Sn/n
所以{Sn/n}是2为公比的等比数列
即通向为s1*2^n-1=2^n-1
an+1=(n+2)*sn/n
所以an=(n+1)2^(n-2)
因为sn/n的通向为2^n-1
所以sn=n*2^n-1