1.
bn=1+(n-1)(4/3)=(4n-1)/3
n=1時(shí),b1=a1/1
a1=b1=1
n≥2時(shí),
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時(shí),a1=2-1=1,同樣滿足通項(xiàng)公式.
a(n+1)-an=2(n+1)-1-2n+1=2,為定值.
數(shù)列{an}是以1為首項(xiàng),2為公差的等差數(shù)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
數(shù)列{cn}的通項(xiàng)公式為cn=n³.