计算:1/(1*3)+1/(2*4)+1/(3*5)+…+1/(n(n+1)(n+2))

Question

过程

Answer 1

最后一项表达式有误，根据前面每项的规律，最后一项应该是：1/[n(n+2)]
而1/[n(n+2)]
=2/[2n(n+2)]
=[(n+2)-n]/[2n(n+2)]
=(n+2)/[2n(n+2)]-n/[2n(n+2)]
=1/2n-1/[2(n+2)]
=[1/n-1/(n+2)]×1/2
则：
1/(1×3)=(1/1-1/3)×1/2
1/(2×4)=(1/2-1/4)×1/2
1/(3×5)=(1/3-1/5)×1/2
……
1/[n(n+2)]=[1/n-1/(n+2)]×1/2
那么：
1/(1×3)+1/(2×4)+1/(3×5)+……+1/[n(n+2)]
=(1/1-1/3)×1/2+(1/2-1/4)×1/2+(1/3-1/5)×1/2+……+[1/n-1/(n+2)]×1/2
=[1/1-1/3+1/2-1/4+1/3-1/5+……+1/n-1/(n+2)]×1/2
=[1/1+1/2-1/(n+1)-1/(n+2)]×1/2
=(3n²+5n)/(4n²+12n+8)

Answer 2

原题应该是算到1/(n(n+2))吧，如果是的话应该利用裂项相消
将1/（n(n+2)）拆成1/2[(1/n)-(1/n+2)]你可以列出前3项再列出后三项~~~~~~~就可以发现规律

Answer 3

写得对吗？应该是1/（1*3）+1/（2*4）1/（3*5）+…+1/(N*（N+2））=1/2（3/2-1/（N+1）-1/（N+2））吧

Answer 4

LZ你题目有没有写错哦？最后一项怎么分母有3项，这跟前面三项不合规律的！叫人怎么做？