等差数列前n项和

题目

等差数列前n项和
an=(1/n+1)+(2/n+1)+...+(n/n+1).又bn=2/an×a(n-1),求bn的前n项和.

答案

∵a[n]=(1/n+1)+(2/n+1)+...+(n/n+1)
=(n+1)/n+(n+2)/n+...+(n+n)/n
=(3n+1)/2
∴a[n+1]-a[n]=3/2 【1】
∵b[n]=2/(a[n]a[n-1])
∴b[n+1]=2/(a[n+1]a[n])=2(1/a[n]-1/a[n+1])/(a[n+1]-a[n])
将【1】式代入上式：
b[n+1]=4(1/a[n]-1/a[n+1])/3
设b[n]的前n项和S[n] (n>1)
∴S[n]=2/(a[2]a[1])+2/(a[3]a[2])+...+2/(a[n]a[n-1]))
=4(1/a[1]-1/a[2])/3+4(1/a[2]-1/a[3])/3+...+4(1/a[n-1]-1/a[n])/3
=4(1/a[1]-1/a[n])/3
∵a[1]=2,a[n]=(3n+1)/2
∴S[n]=4(1/2-2/(3n+1)])/3=4{3(n-1)/[2(3n+1)]}/3=2(n-1)/(3n+1)
即：b[n]的前n项和是：2(n-1)/(3n+1) (n>1)