Question

已知公差不为0的等差数列an的前n项和为Sn,S3=a4+6{Sn/1}Tn,且a1,a4,a13成等比数列.求数列an的通项公式.

求数列{Sn/1}的前n项和Tn的表达式.
是 S3=a4+6{1/Sn}Tn!!!!! 抱歉 写错了 ..

Answer 1

a(n)=a+(n-1)d, d不为0。
s(n) = na+n(n-1)d/2.

s(3)=3a+3d=a(4)+6=a+3d+6,a=3.
a(n)=3+(n-1)d.

[a(4)]^2 = (3+3d)^2= a(1)a(13)=3(3+12d)=9(1+4d),
9(1+d)^2 = 9(1+4d),
(1+d)^2 = 1+2d+d^2 = 1 + 4d, 0 = d^2 -2d=d(d-2). d=2.
a(n) = 3+2(n-1)=2n+1.

s(n)=3n+n(n-1)=n(n+2),
1/s(n) = 1/[n(n+2)] = (n+1)/[n(n+1)(n+2)] = (n+2-1)/[n(n+1)(n+2)] = 1/[n(n+1)] - 1/[n(n+1)(n+2)]
= 1/n - 1/(n+1) - (1/2){ 1/[n(n+1)] - 1/[(n+1)(n+2)] },

t(n) = 1/s(1) + 1/s(2) + 1/s(3) + ... + 1/s(n-1) + 1/s(n)
=1/1-1/2 + 1/2-1/3 + 1/3-1/4 + ... + 1/(n-1)-1/n + 1/n-1/(n+1) -(1/2){1/(1*2)-1/(2*3) + 1/(2*3)-1/(3*4) + 1/(3*4)-1/(4*5) + ... + 1/[(n-1)n]-1/[n(n+1)] + 1/[n(n+1)] - 1/[(n+1)(n+2)] }
= 1/1 - 1/(n+1) - (1/2){1/(1*2) - 1/[(n+1)(n+2)]}
= 3/4 - 1/(n+1) + 1/[2(n+1)(n+2)]
= 3/4 - [2(n+2)-1]/[2(n+1)(n+2)]
= 3/4 - (2n+3)/[2(n+1)(n+2)]
= 3/4 - 1/[2(n+1)] - 1/[2(n+2)]