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
来源:学生作业帮 编辑:神马作文网作业帮 分类:数学作业 时间:2024/10/02 06:08:09
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.
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.
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+1)(n+2)
=(n^2+3n)/4(n+1)(n+2)
=n(n+3)/[4(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+1)(n+2)
=(n^2+3n)/4(n+1)(n+2)
=n(n+3)/[4(n+1)(n+2)]
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