(17)
(1)
等差数列{an} 的前n 项和为 Sn
a3=10, a2,a4,a7 成等比数列
设
an =a1+(n-1)d
a3=10
a1+2d=10 (1)
a2,a4,a7 成等比数列
推导出
a2.a7= (a4)^2
(a1+d)(a1+6d)=(a1+3d)^2
(a1+2d-d)(a1+2d+4d)=(a1+2d+d)^2
(10-d)(10+4d)=(10+d)^2
100+30d-4d^2=100+20d+d^2
5d^2-10d=0
d^2-2d=0
d=2
由(1)式
a1+2d=10
a1+4=10
a1=6
得出
an=a1+(n-1)d = 6+2(n-1) =2n+4
Sn
=a1+a2+...+an
=n(a1+an)/2
=n(6+2n+4)/2
=n(n+5)
(2)
记 bn= 2/(Sn+6) , 求数列 {bn} 的前n项和 Tn
bn
=2/(Sn+6)
=2/(n^2+5n+6)
=2/[(n+2)(n+3)]
=2[ 1/(n+2) -1/(n+3)]
Tn
=b1+b2+...+bn
=2[1/3 -1/(n+3)]