Question

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,

Answer 1

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