数列求和.已知an=（0.5）^n,设bn/an=（2n＋5）/（（2n+1）（2n+3））,求bn的前n项和.

bn/an = (2n+5)/[(2n+1)(2n+3)]= 2/(2n+1) - 1/(2n+3)bn =[2/(2n+1) - 1/(2n+3)] (0.5)^nSn = b1+b2+..+bn= [( 1/3 -1/10)+ (1/10-1/28)+(1/28-1/72)+.( 2/(2n+1)(0.5)^n - 1/[2^n(2n+3)]) ]= 1/3 - 1/[2^n(2n+3)]