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!
2019-05-30
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!
优质解答
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))
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))