数列an满足a1+a2+a3+...+an=n^2,若bn=1/an(an+1),求bn的和sn

因为S(an)=a1+a2+...+an=n^2
所以an = S(an)-S(a(n-1)) = n^2 - (n-1)^2 = 2n-1
因此
bn = 1/ana(n+1) = 1/(2n-1)(2n+1) = 1/2 * (1/(2n-1) - 1/(2n+1))
所以
Sn = b1 + b2 + ...+ bn
= 1/2 * (1 - 1/3 + 1/2 - 1/4 + 1/3 - 1/5 + ...+ 1/(2n-1) - 1/(2n+1))
= 1/2 * (1 + 1/2 - 1/2n - 1/(2n+1))
= 3/4 - 1/4n - 1/(4n+2)
希望有用.

b1=1/a1-1/a2,b2=1/a2-1/a3,…bn=1/an-1/an+1,sn=1/a1-1/an+1。利用a的前n+1项的和减去前n项的和可得到an+1，