Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))求和,不要数学归纳法

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Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))求和,不要数学归纳法
Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))
求和,不要数学归纳法

Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))求和,不要数学归纳法
解.裂项法.
Sn=1/[n(n+1)(n+2)]=(1/2){1/[n)n+1)]-1/[(n+1)(n+2)]}
=(1/2)[1/n-1/(n+1)-1/(n+1)+1/(n+2)]
=(1/2)[1/n-2/(n+1)+1/(n+2)]
=1/1x2x3+1/2x3x4+1/3x4x5+...+1x/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+/4-2/5+1/6
+.+1/n-2/(n+1)+1/(n+2)]
=(1/2)[1-1/2-1/(n+1)+1/(n+2)]
=(n^2+3n)/[4(n+1)(n+2)]
其他类似的题目比如：
1/1x2x3+1/2x3x4+1/3x4x5+------+1/98x99x100
=(1/2)*(1/1*2-1/2*3)+(1/2)*(1/2*3-1/3*4)+...+(1/2)(1/98*99-1/99*100)
=(1/2)*(1/1*2-1/2*2+1/2*3-1/3*4+...+1/98*99-1/99*100)
=(1/2)*(1/2-1/9900)
=(1/2)*(4949/9900)
=4949/19800.

求和Sn=1x2x3+2x3x4+……+n(n+1)(n+2) Sn=1/(1x2x3)+1/(2x3x4)+...+1/(n(n+1)(n+2))求和,不要数学归纳法 1x2x3+2x3x4+...+nxx=? 1x2x3,x100 1x2x3+2x3x4+.+nx(n+1)(n+2)=? 1x2x3+2x3x4+.+n(n+1)(n+2)=? 1x2x3+2x3x4+3x4x5+...+7x8x9=, 1x2X3+2x3X4+3x4X5+…+7X8X9=? 1x2x3……x100= 编程实现:1x2x3...x100 1x2=1|3(1x2x3-0x1x2) 2x3=1|3(2x3x4-1x2x3) 3x4=1|3(3x4x5-2x3x4)由此推断1x2x3+2x3x4+3x4x5.7x8x9=? 1x2=1/3x(1x2x3-0x1x2)2x3=1/3x(2x3x4-1x2x3)3x4=1/3x(3x4x5-2x3x4)求1x2x3+2x3x4+3x4x5+...+7x8x9=? 1/1x2x3+1/2x3x4+.+1/98x99x100 . 1/1X2X3+1/2X3x4+…+1/48X49X50 1X2X3+2X3X4+.+n(n+1)(n+2) 求和1x2x3+2x3x4+...+n(n+1)(n+2) 1x2x3+2x3x4+…+n(n+1) 1x2x3······x100=