n=1时,a1=S1=2×1²+1=3
n≥2时,Sn=2n²+n S(n-1)=2(n-1)²+(n-1)
an=Sn-S(n-1)=2n²+n-2(n-1)²-(n-1)=4n-1
n=1时,a1=4-1=3,同样满足.
数列{an}的通项公式为an=4n-1
1/[ak·a(k+1)]=1/[(4n-1)(4(n+1)-1]=(1/4)[1/(4n-1) -1/(4(n+1)-1]
1/(a1·a2)+1/(a2·a3)+...+1/[an·a(n+1)]
=(1/4)[1/(4×1-1)-1/(4×2-1)+1/(4×2-1)-1/(4×3-1)+...+1/(4n-1)-1/(4(n+1)-1)]
=(1/4)[1/3 -1/(4n+3)
=4/3 -1/[4(4n+3)]
n->+∞,4n+3->+∞ 4(4n+3)->+∞,1/[4(4n+3)]->0
4/3 -1/[4(4n+3)]->4/3
lim[1/(a1·a2)+1/(a2·a3)+...+1/[an·a(n+1)]] =4/3