等比数列﹛an﹜中,a1=a,前n项和为sn,若﹛an+1﹜成等差数列,求sn?

题目

等比数列﹛an﹜中,a1=a,前n项和为sn,若﹛an+1﹜成等差数列,求sn?
需要详解

答案

a1=a,等比=d
b1=a+1,等差=k
a2=a*d
b2=a*d+1=a+1+k
a3=a*d*d
b3=a*d*d+1=a+1+2k
a*d*d=a+2k;
a*d=a+k
(a+k)^2=a^2+2ka+k^2=(a+2k)a
k=0,
d=1
sn=n*a