大师作文网用裂项法求值1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)1/1x3 +
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用裂项法求值
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
=[和1/n(n+1)-和1/(n+1)(n+2)]/2
=[1-1/(n+1)-1/2+1/(n+2)]/2
=1/4-1/2(n+1)(n+2)
1/n(n+2)=[1/n-1/(n+2)]/2
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-1/2(n+1)-1/2(n+2)
1/1x2x3 + 1/2x3x4 +1/3x4x5 + … + 1/n(n+1)(n+2)
=[和1/n(n+1)-和1/(n+1)(n+2)]/2
=[1-1/(n+1)-1/2+1/(n+2)]/2
=1/4-1/2(n+1)(n+2)
1/n(n+2)=[1/n-1/(n+2)]/2
1/1x3 + 1/2x4 + 1/3x5 … + 1/n(n+2)
=[1+1/2-1/(n+1)-1/(n+2)]/2
=3/4-1/2(n+1)-1/2(n+2)
1x2x3+2x3x4+3x4x5+4x5x6+...+n(n+1)(n+2)=
1/(1x2x3)+1/(2x3x4)+1/(3x4x5)+.1/(nx(n+1)x(n+2)=?
1x2X3+2x3X4+3x4X5+…+7X8X9=?
1x2x3+2x3x4+3x4x5+…+8x9x10
求和1x2x3+2x3x4+...+n(n+1)(n+2)
1/1x2x3+1/2x3x4+1/3x4x5
1x2x3+2x3x4+3x4x5+...+7x8x9=,
1x2x3+2x3x4+3x4x5+.+10x11x12
1x2x3+2x3x4+3x4x5……+10x11x12巧算 请在两天之内回答
1x2=1/3(1x2x3=0x1x2 ) 2x3=1/3(2x3x4-1x2x3) 3x4=1/3(3x4x5- 2x
1x2=1|3(1x2x3-0x1x2) 2x3=1|3(2x3x4-1x2x3) 3x4=1|3(3x4x5-2x3x
1x2=三分之一｛1x2x3-0x1x2｝；2x3-三分之一｛2x3x4-1x2x3｝：3x4-三分之一｛3x4x5-2