Question

等差数列an中,a2+a3=7,a4+a5+a6=18,第一问求an的通项公式 ;第二问 设数列an的前n项和为Sn,求

Answer 1

1)设首项为a1,公差为d

a2+a3=7=2a1+3d,

a4+a5+a6=18=3a1+12d

得 a1=2 ,d=1

an=1+n (n为大于0的整数）

2) Sn=na1+n(n-1)d/2

=2n+n^2/2-n/2

=n(n+3)/2

3)1/Sn=2/[n(n+3)]

=2/3[1/n-1/(n+3)]

所以S3+S6+.+S3n

=2/3[1/3-1/6+1/6-1/9+.+1/3(n-1)-1/3n]

= 2[1/3-1/3n]/3

=2(n-1)/9n