Question

已知数列{an}是各项均不为0的等差数列,前n项和为Sn,且(an)^2=S2n-1,求an

Answer 1

数列{an}是各项均不为0的等差数列,设首相a1,公差d

an=a1+（n-1）d,

S2n-1=[a1+（n-1）d]·(2n-1)

(an)^2=S2n-1 → [a1+（n-1）d]²=[a1+（n-1）d]·(2n-1)

数列{an}各项均不为0

所以an=a1+（n-1）d=2n-1

an=2n-1,首项1,公差2