序列真题训练

a1=S1=3

an=2(n-1)+a1

定义{an}是一个等差数列，第一项为3，容差为2。

100 & lt；an = 2n+1 & lt；200，所以50≤n≤99

因此，100

问题2:根据第一个问题，Sn = n 2+2n = n (n+2)。

1/s 1 = 1/1 *(1+2)= 1/2(1/1-1/3)

同样的1/S2 = 1/2(1/2-2/4)。

1/S3 = 1/2(1/3/-3/5)，1/sn = 1/21/n-1/(n+2)

S=1/s1+1/s2+....+1/序列号

=1/2(1/1-1/3)+1/2(1/2-2/4)+1/2(1/3/-3/5)....+1/21/n-1/(n+2)

= 1/2 * 1+1/2 * 1/2-1/2 * 1/(n+1)-1/2 * 1/(n+2)

= 3/4-1/2 * 1/(n+1)-1/2 * 1/(n+2)

= 3/4-1/2(n+1)-1/2(n+1)

第三个问题

an = 1/n+1+1/n+2+...+n/n+1 = n(n+1)/2/n+1 = n/2

bn = 2/an * a(n+1)= 8/n(n+1)= 81/n-1/(n+1)

bn前n项之和为Sn = 81-1/2+1/2-1/3+1/4。。。。。+1/(n-1)-1/n+1/n-1/(n+1)

=81-1/(n+1)

=8-8/(n+1)