Question

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

Answer 1

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