Question

数列an ,a1=1,当n>=2时,an=（根号sn+根号sn-1）/2,证明根号sn是等差数列,求an

Answer 1

由于

an=sn-sn-1=(根号sn)^2-(根号sn-1)^2=（根号sn-根号sn-1）*（根号sn+根号sn-1）

=根号sn+根号sn-1）/2

上面等号两边同时约去（根号sn+根号sn-1）

可得（根号sn-根号sn-1）=1/2

所以根号sn是等差数列.

且由等差数列公式得

根号sn=1+0.5(n-1)=0.5(n+1)

所以sn=(1/4)(n+1)^2,sn-1=(1/4)n^2

所以an=sn-sn-1=(1/4)(2n+1),n>1

=1 ,n=1