数列an ,a1=1,前n项和为Sn ,正整数n对应的n an Sn 成等差数列.1.证明{Sn+n+2}成等比数列,
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数列an ,a1=1,前n项和为Sn ,正整数n对应的n an Sn 成等差数列.1.证明{Sn+n+2}成等比数列,
2.求{n+2/n(n+1)(1+an)}前n项和
2.求{n+2/n(n+1)(1+an)}前n项和
![数列an ,a1=1,前n项和为Sn ,正整数n对应的n an Sn 成等差数列.1.证明{Sn+n+2}成等比数列,](/uploads/image/z/18819443-11-3.jpg?t=%E6%95%B0%E5%88%97an+%2Ca1%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn+%2C%E6%AD%A3%E6%95%B4%E6%95%B0n%E5%AF%B9%E5%BA%94%E7%9A%84n+an+Sn+%E6%88%90%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.1.%E8%AF%81%E6%98%8E%7BSn%2Bn%2B2%7D%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C)
n an Sn 成等差数列 2an=n+Sn 2(Sn-Sn-1)=n+Sn Sn =2 S(n-1)+n
S(n+1)= 2Sn+n+1 S(n+1)+n+1+2= 2( Sn+n+2) ( S(n+1)+n+1+2)/( Sn+n+2) =2
故 {Sn+n+2}成等比数列
再问: 第2问怎么做
再答: {Sn+n+2}成等比数列 求出Sn an=Sn-Sn-1 代如整理在求和
再问: ?整理不出来
再答: {Sn+n+2}成等比数列 Sn=4*2^(n-1)=2*2^n -n-2 2an=n+Sn=2*2^n -2 an=2^n -1 代入后是否可求 你试一试
S(n+1)= 2Sn+n+1 S(n+1)+n+1+2= 2( Sn+n+2) ( S(n+1)+n+1+2)/( Sn+n+2) =2
故 {Sn+n+2}成等比数列
再问: 第2问怎么做
再答: {Sn+n+2}成等比数列 求出Sn an=Sn-Sn-1 代如整理在求和
再问: ?整理不出来
再答: {Sn+n+2}成等比数列 Sn=4*2^(n-1)=2*2^n -n-2 2an=n+Sn=2*2^n -2 an=2^n -1 代入后是否可求 你试一试
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