题目

用数学裂项法证明：1/1*2*3+1/2*3*4+...+1/N(N+1)(N+2)=N(N+3)/4(N+1)(N+2)
证：1/[N(N+1)(N+2)]=...=1/2[1/N-2/(N+1)+1/(N+2)]
S=1/1*2*3+1/2*3*4+1/3*4*5+...+1/N(N+1)(N+2)
=1/2[1/1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+1/4-2/5+1/6+...+1/N-2/(N+1)+1/(N+2)]
是如何展开运算=1/2[1-1/2-1/(N+1)+1/(N+2)]=（N^2+3N）/[4(N+1)(N+2)]的?请不省步骤展开说明.

提问时间：2020-08-02

答案

1/1*2*3+1/2*3*4+……1/n(n+1)(n+2)=1/2(1/1*2-1/2*3)+1/2(1/2*3-1/3*4)+...+1/2[1/n(n+1)-1/(n+1)(n+2)]=1/2[1/1*2-1/2*3+1/2*3-1/3*4+...+1/n(n+1)-1/(n+1)(n+2)]=1/2[1/1*2-1/(n+1)(n+2)]=1/2*[(n+1)(n+2)-2]/2(n...