Question

已知数列﹛an﹜是公差为2,首项a1=1的等差数列,求数列﹛2^an﹜的前n项和sn

Answer 1

a(n+1)-an=2

2^a(n-1)÷2^an

=2^[a(n+1)-an]

=2²

=4

所以是等比数列,q=4

2^a1=2

所以Sn=2*(1-4^n)/(1-4)=2(4^n-1)/3