1/[n(n+2)]={(1/n)-[1/(n+2)]}/2
所以原式等于0.5×(1/1-1/3+1/2-1/4+1/3-1/5+……+1/99-1/101)
=0.5×[(1/1+1/2+1/3+……+1/99)-(1/3+1/4+1/5+……+1/101)]
=0.5×(1+1/2-1/100-1/101)=14949/20200
1楼中间相多去了几项
1/[n(n+2)]={(1/n)-[1/(n+2)]}/2
所以原式等于0.5×(1/1-1/3+1/2-1/4+1/3-1/5+……+1/99-1/101)
=0.5×[(1/1+1/2+1/3+……+1/99)-(1/3+1/4+1/5+……+1/101)]
=0.5×(1+1/2-1/100-1/101)=14949/20200
1楼中间相多去了几项