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裂项法数列求和的一些题目,

来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/16 06:18:46
裂项法数列求和的一些题目,
这个..
n(n+1)=[n(n+1)(n+2)-(n-1)n(n+1)]/3
1*2+2*3+...+(n-1)n+n(n+1)=(1/3)[1*2*3-0*1*2 + 2*3*4-1*2*3 +...+(n-1)n(n+1)-(n-2)(n-1)n +n(n+1)(n+2)-(n-1)n(n+1)] = (1/3)[n(n+1)(n+2)]=n(n+1)(n+2)/3
1*2+2*3+...+50*51=50*51*52/3=50*17*2*26=44200
n(n+1)(n+2)=[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]/4
1*2*3+2*3*4+...+(n-1)n(n+1)+n(n+1)(n+2)=(1/4)[1*2*3*4-0*1*2*3 + 2*3*4*5-1*2*3*4 +...+(n-1)n(n+1)(n+2)-(n-2)(n-1)n(n+1) + n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
=(1/4)[n(n+1)(n+2)(n+3)]
=n(n+1)(n+2)(n+3)/4
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1+2+...+n=n(n+1)/2
1/[1+2+...+n]=2/[n(n+1)] = 2/n - 2/(n+1)
1/[1+2] + 1/[1+2+3] + ...+ 1/[1+2+...+(n-1)] + 1/[1+2+...+n]
=2/2 - 2/3 + 2/3 - 2/4 + ...+2/(n-1) - 2/(n) + 2/n - 2/(n+1)
=1 - 2/(n+1)
=(n-1)/(n+1)