Question

已知数列﹛an﹜是公差为2的等差数列。其前N项和为Sn，且a1，a4,a16成等比数列

（1）求数列﹛an﹜的通项公式﹙2﹚求﹛1／Sn﹜的前n项和Tn

Answer 1

1,令an=a1+(n-1)d
d=2
an=a1+2(n-1)
a4=a1+3--------(1)
a16=a1+30--------(2)
a4^2=a1*a16-----------(3)
由(1)(2)(3)得a1=2
an=2n
2,sn=n*(n+1)
1/sn=1/n*(n+1)
Tn=s1+s2+s3+......sn
=1/(1*2)+1/(2*3)+1/(3*4)+..............+1/n*(n+1)
=1/1-1/2+1/2-1/3+1/3-1/4+...............+1/n-1/(1+n)
=1-1/(1+n)

Answer 2

(1)a1，a4,a16成等比数列
a4^2 = a1a16
(a1+3x2)^2 = a1(a1+15x2)
a1^2+12a1+36 = a1^2+30a1
18a1 = 36
a1 = 2
an = 2n
(2)Sn = (2+2n)n/2 = n(n+1)
Tn = 1/1x2+1/2x3+...+1/nx(n+1)
=1-1/2+1/2-1/3+...+1/n-1/n+1
=n/(n+1)
你们考试吗？刚才有人问了一个这个

Answer 3

(1)a1，a4,a16成等比数列
a4^2 = a1a16
(a1+3x2)^2 = a1(a1+15x2)
a1^2+12a1+36 = a1^2+30a1
18a1 = 36
a1 = 2
an = 2n
(2)Sn = (2+2n)n/2 = n(n+1)
Tn = 1/1x2+1/2x3+...+1/nx(n+1)
=1-1/2+1/2-1/3+...+1/n-1/n+1
=n/(n+1)