Question

已知公差不为0的等差数列﹛a﹜，其前n项和为Sn，S3=a4+6,且a1,a4,a13成等比数列。求数列﹛an﹜的通项公式

并求数列﹛1／Sn﹜的前n项和Tn

Answer 1

S3=3a1 + 3*(3-1)/2d=3a1+3d
a4=a1 + 3d
则 3a1 + 3d=a1 +3d+6 解得 a1=3
(a4)^2=a13 * a1 即 (3+3d)^2 = (3+12d)*3
解得 d=2 或 d=0(舍去)
所以 an = 3+2n Sn=3n+n(n-1)*2/2=n(n+2)
所以 1/Sn=1/(n(n+2))=(1/n-1/(n+2))*1/2
所以 Tn=1/S1+1/S2+1/S3+1/S4+。。。。。。+1/Sn
=1/2*(1-1/3)+1/2*(1/2-1/4)+1/2*(1/3-1/5)+1/2*(1/4-1/6)+。。。。。。+1/2*(1/n-1/(n+2))
=1/2*(1-1/3+1/2-1/4+1/3-1/5+1/4-1/6。。。。。。+1/n-1/(n+2))
=1/2*(1+1/2-1/(n+1)-1/(n+2))
=3/4-(2n+3)/(2(n+1)(n+2))

Answer 2

设公差为x
则：a1+a1+x+a1+2x=a1+6 (a1+a2+a3=s3=a4+6)
解：a1=3
由a1/a4=a4/a13 a4a4=a1a13
带入a1=3
解x=2