数列:裂开成差

来源：百度知道 编辑：UC知道 时间：2024/05/27 18:06:08
如下题:1/(1*3*5)+2/(3*5*7)+...+1/(95*97*99)=?

老师说要用 裂开成差 .请解题并7.

裂项相消法：
1/(1*3*5)+1/(3*5*7)+...+1/(95*97*99)
=1/4*[1/(1*3)-1/(3*5)]+1/4*[1/(3*5)-1/(5*7)]+…+1/4*[1/(95*97)-1/(97*99)]
=1/4*[1/(1*3)-1/(3*5)+1/(3*5)-1/(5*7)+…+1/(95*97)-1/(97*99)]
=1/4*[1/(1*3)-1/(97*99)]
=1/4*(1/3-1/9603)
=1/4*3200/9603
=800/9603
附公式：
1/n(n+a)(n+2a)=1/(2a)*[1/n(n+a)-1/(n+a)(n+2a)]

1/(1*3*5)=(1/4)*(1/(1*3)-1/(3*5))
同理其他也可这么展开
故1/(1*3*5)+1/(3*5*7)+...+1/(95*97*99)=(1/4)(1/(1*3)-1/(3*5)+1/(3*5)-1/(5*7)+…+1/(95*97)-1/(97*99))=（1/4)(1/(1*3)-1/(97*99))