求一道数列题：an=3/(2n+3)(4n+10) 用裂项相消法做.（尽量详细些,我比较笨

an =(3/4)* [(4n+10)-2*(2n+3)]/[(2n+3)(4n+10)]=(3/4)[1/(2n+3)-1/(2n+5)]an-1=(3/4)[1/(2n+1)-1/(2n+3)]an-2=(3/4)[1/(2n-1)-1/(2n+1)]..a1=(3/4)(1/5-1/7)Sn=(3/4)[1/5-1/(2n+5)]