Question

等差数列{an}的公差为d,bn=(a1+a2+…+an)/n,求证:数列{bn}是等差数列

Answer 1

a1+a2+…+an=[a1+(n-1)×d]×n÷2
bn=[a1+(n-1)×d]÷2=½a1+½(n-1)d
即以½a1为首项½d为公差的等差数列

Answer 2

bn=(a1+an)×n/2n=(a1+an)/2=[a1+a1+(n-1)d]/2=a1+1/2(n-1)d
∴bn为公差为1/2d的等差数列

Answer 3

bn-b(n-1)=Sn/n-S(n-1)/n-1
=a1+(n-1)d/2-a1-(n-2)d/2
=d/2
d/2为一个常数，所以{bn}是公差为d/2的等差数列

Answer 4

会~
因为a1+....+an=3n^2,所以首项a1为3，公差为6（因为a2=s2-a1=12-3=9)
所以an=6n-3咯
bn就是9+21+33+.....+(12n-3)=6n^2+3n(等差数列求和）
cn=3+15+27+.....+(12n-9)=6n^2-3n
ok~