Question

已知数列an是公差不为零的等差数列,其前n项和为Sn,若S1+S3=7/2,a4,a8,a16依次等比数列,求证Sn=a

Answer 1

设首项为a1,公差为d,则有：S1+S3=a1+a1+a2+a3=4a1+3d=7/2

∵a4,a8,a16依次等比数列

∴a8²=a4a16即（a1+7d）²=（a1+3d）（a1+15d）化简得a1d-d²=0

∵公差d≠0

∴a1-d=0即a1=d,代入4a1+3d=7/2解得a1=d=1/2

∴an=1/2+（n-1）/2=n/2 a(n+1)=(n+1)/2 ana(n+1)=n(n+1)/4=(n²+n)/4

Sn=n/2+n(n-1)/2×1/2=(n²+n)/4,得证