bn}是首项为1,公差4/3的等差数列,且bn=（a1+2a2+……+nan）/（1+2+……+n）, 1.求证{an}为等差数列

1.
bn=1+(n-1)(4/3)=(4n-1)/3
n=1时,b1=a1/1
a1=b1=1
n≥2时,
bn=(a1+2a2+...+nan)/(1+2+...+n)=(a1+2a2+...+nan)/[n(n+1)/2]
[n(n+1)/2]bn=a1+2a2+...+nan (1)
[n(n-1)/2]b(n-1)=a1+2a2+...+(n-1)a(n-1)(2)
(1)-(2)
[n(n+1)/2]bn -[n(n-1)/2]b(n-1)=nan
an=[(n+1)/2]bn -[(n-1)/2]b(n-1)
=[(n+1)/2][(4n-1)/3]-[(n-1)/2][(4n-5)/3]
=2n-1
n=1时,a1=2-1=1,同样满足通项公式.
a(n+1)-an=2(n+1)-1-2n+1=2,为定值.
数列{an}是以1为首项,2为公差的等差数列.
2.
cn=a[n(n-1)/2 +1]+a[n(n-1)/2 +2]+...+[n(n-1)/2 +n]
=na[n(n-1)/2] +(1×2+2×2+...+n×2)
=n×[2n(n-1)/2 -1]+2(1+2+...+n)
=n³
数列{cn}的通项公式为cn=n³.