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,
问题描述：
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
答
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}成等比数列第2问怎么做{Sn+n+2}成等比数列 求出Snan=Sn-Sn-1 代如整理在求和？整理不出来{Sn+n+2}成等比数列 Sn=4*2^(n-1)=2*2^n -n-2 2an=n+Sn=2*2^n -2an=2^n -1 代入后是否可求 你试一试