www.5129.net > 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9=

1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9=

n=10 ,很好算得,得440。也许我的方法很麻烦,但是思路是很不错的,可以1x2 2x3 3x4 4x5 5x6 6x7 7x8 8x9 9x10 10x11 =2 6 12 20

因为an=n*(n-1)/n+1 所以每个数可以看成2/3+(1+2/4)+(2+2/5)+(3+2/6)...+(8+2/11)= 1+2+3...+8+2/3+2/4+2/5...+2/11=36+2*(1/3+1/4+1/5+...1/11) 因为后面的那个是一个发散数列,好像没有求和公式,所以只能同分计算,当然可以把简单...

1/4X5 + 1/5X6 + 1/6X7 + 1/7X8 + 1/8X9 + 1/9X10 = (1/4 - 1/5) + (1/5 - 1/6) + (1/6 - 1/7) + (1/7 - 1/8) + (1/8 -1/9) + (1/9 - 1/10) = 1/4 - 1/10 = 3/20

y=5x6+6x7+7x8+8x9+9x10+10x11+11x12 y=5*6+6*7+7*8+8*9+9*10+10*11+11*12 = 532

1/[n×(n+1)]=1/n-1/(n+1) ∴原式=(1/3-1/4)+(1/4-1/5)+(1/5-1/6)+(1/6-1/7)+(1/7-1/8)+(1/8-1/9) =1/3-1/9 =2/9

(1十2X3十4X5十6)X7十8X9二303

1十2x3十4x5十6x7十8x9=303在算是中添加合适的括号: 这样算: (1+2x3+4x5+6)x7+8x9=303 ,

1x2 + 2x3 + 3x4 + ...... + n(n+1) = 1/3 * n(n+1)(n+2) 1x2+2x3+3x4+4x5+5x6+6x7+7x8+8x9 = 1/3 * 8 * 9 * 10 = 240

先求它们的倒数和,再把结果倒回来。

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