Question

已知数列{an}的通项公式an=2n+1，由bn=a1+a2+a3+…an/n所确定的数列{bn}的前n项之和是

Answer 1

数列{an}的前n项和Sn=(a1+an)n/2=(n+2)n
所以bn=Sn/n=n+2为等差数列
故数列{bn}的前n项之和Tn=(b1+bn)n/2=(3+n+2)n/2=(n^2+5n)/2

如果是岁缺bn=a1+a2+a3+…咐悔+（an/n）
=（Sn-1）+an/n
=[(n+1)(n-1)]+(2n+1)/n
=n^2-1+2+1/n
=n^2+(1/衡雀正n)+1
故数列{bn}的前n项之和Tn=(1^2+2^2+3^2+ +n^2)+(1/1+1/2+1/3+ +1/n)+n

Answer 2

an=2n+1
bn=(a1+a2+...+an)/n
=[2(1+2+...+n)+n]/早前n
=[n(n+1)+n]/肆凯n
=n+2
Sn=b1+b2+...+bn
=(1+2+...+n)+2n
=n(n+1)/2+2n
=n(n+5)/裂睁唤2

Answer 3

bn=a1+a2+a3+…an/n??????????