数列An的前n项和Sn,A(1)=1,A(n+1)=(n+2)Sn/n,证明1.Sn/n是等差数列 2.S(n+1)=4An
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![数列An的前n项和Sn,A(1)=1,A(n+1)=(n+2)Sn/n,证明1.Sn/n是等差数列 2.S(n+1)=4An](/uploads/image/z/10213847-71-7.jpg?t=%E6%95%B0%E5%88%97An%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%2CA%281%29%3D1%2CA%28n%2B1%29%3D%EF%BC%88n%2B2%29Sn%2Fn%2C%E8%AF%81%E6%98%8E1.Sn%2Fn%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+2.S%28n%2B1%29%3D4An)
数列An的前n项和Sn,A(1)=1,A(n+1)=(n+2)Sn/n,证明1.Sn/n是等差数列 2.S(n+1)=4An
数列An的前n项和Sn,A(1)=1,A(n+1)=(n+2)Sn/n,证明1.Sn/n是等差数列 2.S(n+1)=4An
数列An的前n项和Sn,A(1)=1,A(n+1)=(n+2)Sn/n,证明1.Sn/n是等差数列 2.S(n+1)=4An
A(n+1)=(n+2)Sn/n
因为S(n+1)-Sn=an
所以S(n+1)-Sn=(n+2)Sn/n
n*S(n+1)-n*Sn=(n+2)*Sn
n*S(n+1)=2*(n+1)Sn
所以S(n+1)/(n+1)=2*Sn/n
又a1=1
所以S1/1=1
所以{Sn/n}是首项为1,公比为2的等比数列
Sn/n=2^(n-1)
Sn=n*2^(n-1)
某不才,只求出它是等比...
不过还是继续吧~
S(n+1)=(n+1)*2^n
又因为an=Sn-S(n-1)=n*2^(n-1)-(n-1)*2^(n-2)=2n*2^(n-2)-(n-1)*2^(n-2)=(n+1)*2^(n-2)
所以4an=(n+1)*2^n=S(n+1)