已知{an}是公差不为0的等差数列,a1=1,且a1,a3,a9成等比数列,求数列{2^an}的前n项和Sn

设公差为d,则d≠0a1,a3,a9成等比数列,则a3²=a1·a9(a1+2d)²=a1(a1+8d)a1=1代入,整理,得d²-a1d=0d(d-a1)=0d≠0,因此只有d-a1=0 d=a1=1an=a1+(n-1)d=1+1×(n-1)=n2^(an)=2ⁿ2^(a1)=2 2^[a(n+1)]/2...