bn=1/(2n+1)(2n+1),球bn的前N项和Tn

问题描述:

bn=1/(2n+1)(2n+1),球bn的前N项和Tn

是bn=1/(2n+1)(2n-1)吧
b1=1/3=1/2*(1- 1/3 ),b2=1/15=1/2*(1/3-1/5),b3=1/35=1/2*(1/5-1/7),……bn=1/(2n+1)(2n-1)=1/2*[1/2n-1)-(1/2n-1)]
Tn=b1+b2=+b3+……+bn
=1/3+1/15+1/35+……+1/(2n+1)(2n-1)
=1/2*(1-1/3)+1/2*(1/3-1/5)+1/2*(1/5-1/7)+……+1/2*[(1/2n-1)-(1/2n+1)]
=1/2*[1- 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + …… +(1/2n-1)-(1/2n+1)]
=1/2*[1-(1/2n+1)]
=1/2*[2n/2n+1)]
=n/2n+1
故Tn=n / 2n+1