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1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+1/(5*6*7)+1/(6*7*...

1只+1只=1双; 1(个月)+2(个月)=1季度 3天+4天=1周 5个月+7个月=1年 6(小时)+18小时=1天

1/[n(n+1)(n+2)]=1/2*[1/n-2/(n+1)+1/(n+2)], 例如 ∴1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+1/(4*5*6)+1/(5*6*7)+1/(6*7*8)+1/(7*8*9) =1/2*[1-2/2+1/3+1/2-2/3+1/4+1/3-2/4+1/5+……+1/7-2/8+1/9] =1/2[1-1/2-1/8+1/9] =1/2(1/2-1/72) =35/144. 后面加一...

1、这是一道六年级的数奥题,要用“分数的拆分法”来解答才行。 2、就是说,把题目中的所有分数拆成两个分数相减的形式,形成加减相互抵消,从而达到不同分的目的。 3、这道题首先把题中的分数拆分成下面的样子: 1/1*2*3=[1/(1×2)-1/(2×3)]×...

原是=4/2(1/1*2-1/2*3)+5/2(1/2*3-1/3*4)+6/2(1/3*4-1/4*5)+…… +11/2(1/8*9-1/9*10) 这是第一次裂项,将(1*2*3)分之4裂成4/2(1/1*2-1/2*3) (2*3*4)分之5裂成5/2(1/2*3-1/3*4)……依次类推 继续运算得 原是=1+1/2{1/2*3+1/3*4+……+1/8*9}-(11/2...

解法一: 1×2+2×3+3×4+...+n(n+1) =⅓×[1×2×3-0×1×2+2×3×4-1×2×3+3×4×5-2×3×4+...+n(n+1)(n+2)-(n-1)n(n+1)] =⅓n(n+1)(n+2) 解法二: 考察一般项第k项,k(k+1)=k²+k 1×2+2×3+3×4+...+n(n+1) =(1²+2²+3²+...+n...

1/1*2*3+1/2*3*4+1/3*4*5+1/4*5*6+1/5*6*7+1/6*7*8+1/7*8*9+1/8*9*10+1/9*10*11 =(1/2)[(1/1*2)-(1/2*3)]+(1/2)[(1/2*3)-(1/3*4)]+(1/2)[(1/3*4)-(1/4*5)]+...+(1/2)[(1/9*10)-(1/10*11)] =(1/2)[(1/1*2)-(1/2*3)+(1/2*3)-(1/3*4)+(1/3*4)-(1/4*...

观察:2*4,3*5……,整个式子中只有它在变,设它为(n-1)*(n+1),因为这样的话,第一项取3,第二项取4…… 通分:得到每一项的通式(n^2-1-3)/(n^2-1)=(n+2)(n-2)/(n+1)(n-1) 【3

解:1/(4×5×6)=1/2[1/(4×5)-1/(5×6)] 1/(5×6×7)=1/2[1/(5×6)- 1/(6×7 )] 1/(6×7×8)=1/2[1/(6×7)- 1/(7×8)] ……… 1/(98×99×100)=1/2[1/(98×99)-1/(99×100)] ∴1/(4×5×6)+ 1/(5×6×7)+ 1/(6×7×8)+…..+ 1/(98×99×100) =1/2[1/(4×5)-1/(5×6)]+ 1/2[1/(5...

可以直接学STM32 不过学好51单片机对学习STM32肯定是有邦助的, 起码一些常见外围器件都知道怎么工作的

1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+···+1/99*100 =(1-1/2)+(1/2-1/3)+(1/3-1/4)...+(1/99-1/100) =1-1/2+1/2-1/3+1/3-1/4+....1/99-1/100 =1-1/100 =99/100

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