已知公差不为0的数列{an}的首相为a1=1,前n项的和为Sn,若数列{Sn⼀an}是等差数列

2025-06-21 13:42:47
推荐回答(1个)
回答1:

S1/a1=1
S2/a2-S1/a1=(2+d)/(1+d)-1=d/(1+d)
S3/a3-S1/a1==(3+3d)/(1+2d)-1=(2+d)/(1+2d)

2*d/(1+d)=(2+d)/(1+2d)
解得d=1,d=0(舍去)
所以,an=n

(b(n+1))^2=q^(n+1)(n+2)=q^(n*n+3n+2)
2*bn*b(n+2)=2*q^{n(n+1)/2+(n+1)(n+2)/2}=2*q^(n*n+3n+3)