Question

数列{an}的通项公式为an=4n-1,令bn=a1+a2+……+an/n,则数列{bn}的前n项和为

Answer 1

bn=[4（n+1）n/2-n]/n=2n+1所以所求为2（1+n）×n/2+n=n^2+2n

Answer 2

先求出an的前n项和
Sn = (a1+an)*n/2 = (4*1-1 +4n-1)*n/2 = (4n+2)*n/2 = 2n² +n

bn = Sn/n = 2n+1
数列{bn}的前n项和
Tn = (b1+bn)*n/2 = (2*1+1 +2n+1)*n/2= (2n+4)*n/2= n(n+1) = n² +2n

Answer 3

an=4n-1
an+1=4(n+1)-1
an+-an=4
a1=4-1=3
故数列{an}是一个公差为4,首项为3的等差数列
bn=a1+a2+.....+an/n=[na1+4n(n-1)/2]/n=3+2n-2=1+2n

Answer 4

解：∵an=4n-1∴d=4，a1=5∴数列{an}的前n项和为2n�0�5+3n∵bn=a1+a2+……+an/n∴bn=2n+3∴bn的前n项和为n�0�5+4n