(1)an-a(n-1)=2n-1-(2(n-1)-1)=2 故an为等差数列
(2)sn=n(a1+an)/2=n^2
(3) bn=sn/n=n 因此 1/b1*b2+1/b2*b3+...+1/bn-1*bn=1/(1*2)+1/(2*3)+.1/((n-1)*n)
=1-1/2+1/2-1/3+.+1/(n-1)-1n
=1-1/n
(1)an-a(n-1)=2n-1-(2(n-1)-1)=2 故an为等差数列
(2)sn=n(a1+an)/2=n^2
(3) bn=sn/n=n 因此 1/b1*b2+1/b2*b3+...+1/bn-1*bn=1/(1*2)+1/(2*3)+.1/((n-1)*n)
=1-1/2+1/2-1/3+.+1/(n-1)-1n
=1-1/n