Question

请教数学题，好的话有加分~~~谢谢~~~

已知{an}是公差不为零的等差数列,a1=1,且a1,a3,a9成等比数列,求1/(a1*a3)+1/(a2*a4）+1/(a3*a5)+…+1/(an*an+2)
an的通项公式是an=n(n>1)
要有详细解答过程
谢谢各位高手帮忙！！！

Answer 1

由a1,a3,a9成等比数列得，(1+2d)^2=1+8d
解得，d=1
所以an=n
1/n*(n+2)=1/2[1/n-1/(n+2)]
1/(a1*a3)+1/(a2*a4）+1/(a3*a5)+…+1/(an*an+2)=1/1*3+1/2*4+1/3*5+...+1/n*(n+2)=1/2[(1-1/3)+(1/2-1/4)+(1/3-1/5)+...(1/n-1/(n+2))=1/2[1+1/2-1/(n+1)-1/(n+2)]=
3/4-1/2(n+1)-1/2(n+2)

Answer 2

1/(an*an+2)=1/(n*(n+2))=0.5*(1/n-1/(n+2))
1/(a1*a3)+1/(a2*a4）+1/(a3*a5)+…+1/(an*an+2)=0.5*(1-1/3+1/2-1/4+...+1/n-1/(n+2))=0.5*(1+1/2-1/(n+1)-1/(n+2))=(3n^2+5n)/(4*(n+1)*(n+2))