数列an,bn满足anbn=1,an=n^2+3n+2,则bn的前n项之和为_数学解答

数列an,bn满足anbn=1,an=n^2+3n+2,则bn的前n项之和为

人气：267 ℃　时间：2019-08-19 16:57:43

解答

an=(n+1)(n+2)anbn=1bn=1/an=1/[(n+1)(n+2)]=[(n+2)-(n+1)]/[(n+1)(n+2)]=(n+2)/[(n+1)(n+2)]-(n+1)/[(n+1)(n+2)]=1/(n+1)-1/(n+2)所以Sn=(1/2-1/3)+(1/3-1/4)+……+[1/(n+1)-1/(n+2)]=1/2-1/(n+2)=n/(2n+4)