数列{An}与{Bn}满足:A1=2,Bn=An+A(n+1),(n∈N*),且{Bn}的前n项和为Sn=1/2*n(n+1)

1.
Sn=1/2×n(n+1)
1/Sn=2/(n(n+1)=2(1/n-1/(n+1))
n→∞
lim(1/S1+1/S2+……+1/Sn)
=lim2(1/1-1/2+1/2-1/3+……+1/n-1/(n+1))
=lim2(1-1/(n+1)
=2

2.
Bn=Sn-Sn-1
=1/2×n(n+1)-1/2×(n-1)n
=n
n=1时，B1=S1=1/2×1×2=1，满足上式
Bn=An+A(n+1)=n
A(n+1)+A(n+2)=n+1
两式相减
A(n+2)-An=1
数列{An}的奇数项和偶数项分别成等差数列，公差为1
A1+A2=1 A2=1-A1=1-2=-1
A(2n-1)=A1+(n-1)×1=n+1
A(2n)=A2+(n-1)×1=n-2
A(2n-1)+A(2n)=2n-1
新数列{A(2n-1)+A(2n)}是两个等差数列相加，首项为1，公差为2
T2n=(1+2n-1)n/2=n^2

no points,no answer