Sn+1/Sn+2=an带入n=2 又an=a1+(n-1)d 故得(3a1+2d)/(4a1+6d)=a1+d
带入a1 即得d Sn=na1+n(n-1)d/2
http://zhidao.baidu.com/question/534128756.html
n>=2时 an=Sn-S(n-1)
又an=Sn+1/Sn+2可得 -S(n-1)=1/Sn+2
-S(n-1)-1=1/Sn+1
-[1+S(n-1)]=(1+Sn)/Sn
对之取倒数得 -1/[1+S(n-1)]=Sn/(1+Sn)=1-1/(1+Sn)
即 1/(1+Sn)-1/[1+S(n-1)]=1
因此 {1/(1+Sn)}是公差为1的等差数列
又S1=a1=-2/3 1/(1+S1)=3 所以等差数列首项为3
则 1/(1+Sn)=3+(n-1)*1=n+2
则 Sn=[1/(n+2)]-1
n=1时 S1=a1=-2/3 符合 S1=[1/(1+2)]-1=-2/3
综合得之 Sn=[1/(n+2)]-1
其中是Sn +1还是S(n+1),即那1,2是属于下标还是?