已知等差数列an=1+2n,数列bn=1/an^2-1,求数列bn前n项和Tn
问题描述:
已知等差数列an=1+2n,数列bn=1/an^2-1,求数列bn前n项和Tn
答
bn=1/((1+2n)^2-1)=1/(4n^2+4n)=1/2n-1/(2n+1)
Tn=1/2-1/(2n+1)=(2n-1)/(4n+2)