取An＝[1+2+3+...+(n+1)]/[2+3+...+(n+1)][n≥1,n∈Z] ,则
An=0.5(n+1)(n+2)/[0.5(n+1)(n+2)-1]=(n+1)(n+2)/n(n+3),
原式＝A1×A2×A3×...×An
=[(2×3)/(1×4)]×[(3×4)/(2×5)]×[(4×5)/(3×6)]×...×[(n+1)(n+2)/n(n+3)][共n个分数]
=(2×3)(3×4)(4×5)×...×[(n+1)(n+2)]/(1×4)(2×5)(3×6)×...×[n(n+3)][分子、分母分别相乘]
={[2×3×4×...×(n+1)][3×4×5×...×(n+2)]}/{[1×2×3×...×n][4×5×6×...×(n+3]}
=3(n+1)/(n+3).
取n=1992,得
原式＝A1×A2×A3×...×A1992=（3*1993）/1995=1993/665.
然后用裂项法约分=3/2*(3*4*..*1993/2*3*4..*1992)=3/2*(1993/2)*(4/1995)=1993/665
(1+2)\/2*(1+2+3)\/(2+3)*(1+2+3+4)\/(2+3+4)*…(1+2+3+…1993)\/(2+3+...
解答：取An＝[1+2+3+...+(n+1)]\/[2+3+...+(n+1)][n≥1,n∈Z] ,则 An=0.5(n+1)(n+2)\/[0.5(n+1)(n+2)-1]=(n+1)(n+2)\/n(n+3),原式＝A1×A2×A3×...×An =[(2×3)\/(1×4)]×[(3×4)\/(2×5)]×[(4×5)\/(3×6)]×...×[(n+1)(n+...
(1+2)\/2 * (1+2+3)\/(2+3) * (1+2+3+4)\/(2+3+4) * (1+2+3+4+---+200...
=（2*3）\/（1*4）*（3*4）\/（2*5）*（4*5）\/（3*6）*（5*6）\/（4*7）*.（2001*2002）\/（2000*2003）=3*2002\/2003 =6006\/2003
(1+2)\/2x(1+2+3)\/(2+3)x(1+2+3+4)\/(2+3+4)x...x
=(n^2+n)\/(n^2+n-2)=n(n+1)\/(n+2)(n-1)所以[(1+2)\/2]*[(1+2+3)\/(2+3)]*[(1+2+3+4)\/(2+3+4)]*...*[(1+2+3+...+50)\/(2+3+...+50)]=(2*3\/1*4)*(3*4\/2*5)(4*5\/3*6)...(49*50\/48*51)(50*51\/49*52)=3*50\/52 =75\/26 ...
(2\/1+2)*(2+3\/1+2+3)*(2+3+4\/1+2+3+4)...(2+3+4+...50\/1+2+3+4...
∵1+2+3+4+5+6+……+n=n(n+1)\/2 ∴2+3+4+5+6+……+n=[n(n+1)\/2]-1 ∴(1+2+3+4+5+6+……+n）\/（2+3+4+5+6+……+n)=n(n+1)\/(n+2)(n-1)即（1+2）\/2=2×3\/(1×4)(1+2+3)\/(2+3)=3×4\/(2×5)(1+2+3+4)\/(2+3+4)=4×5\/(3×6)....
...+3\/[(1+2)*(1+2+3)]+4\/[(1+2+3)*(1+2+3+4)]+...+100\/[(1+2+...
1\/[n*(n+1)]=1\/n-1\/(n+1)1\/[n*(n+2)]=1\/2{1\/n-1\/(n+2)} 即 1\/[n*(n+m)]=1\/m{1\/n-1\/(n+m)} 你可以自己代数字 试一试 所以 该题 原式=1\/1-1\/3+1\/3-1\/6+...+1\/(99*100\/2)-1\/(100*101\/2)=1-1\/5050=5049\/5050 ...
...1×(1+2)\/2+(1+2)×(1+2+3)\/3+(1+2+3)×(1+2+3+4)+…+(1+2+3+...
原式=4×（1\/1×2×3+1\/2×3×4+……+1\/99×100×101）=4×1\/2×（1\/1×2-1\/2×3+1\/2×3-1\/3×4+……+1\/99×100-1\/100×101）=2×（1\/2-1\/10100）=1-1\/5050 =5049\/5050
1+2\/2*1+2+3\/2+3*1+2+3+4\/2+3+4*……*1+2……+2001\/2+3+…
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1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+4+...+1999) 简便
解题思路：1+2=2*3\/2 1\/(1+2)=2\/(2*3)=2*(1\/2-1\/3)1+2+3=3*4\/2 1\/(1+2+3)=2\/(3*4)=2*(1\/3-1\/4)………1+2+3+……+1999=1999*2000\/2 1\/(1+2+3+……+1999)=2*(1\/1999-1\/2000)解：1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+....
...1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)...1\/(1+2+3+4...2004)
1\/(1+2+3+...+n)=1\/[n(n+1)\/2]=2\/[n(n+1)]=2[1\/n-1\/(n+1)]1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)...1\/(1+2+3+4...2004)=2(1\/2-1\/3+1\/3-1\/4+...+1\/2004-1\/2005)=2(1\/2-1\/2005)=1-2\/2005 =2003\/2005 ...
(1*1+2*2)\/(1*2)+(2*2+3*3)\/(2*3)+(3*3+4*4)\/(3*4)...+(2000*2000+200...
这样的题一般是要对每一项进行拆分 抽象出来应该是:(n^2+(n+1)^2)\/(n*(n+1))=n\/(n+1) + (n+1)\/n =1-1\/(n+1) + 1 + 1\/n =2 + 1\/n - 1\/(n+1)所以原式= 2*2000 + 1\/1-1\/2 + 1\/2 -1\/3 + ……+ 1\/2000 - 1\/2001 =4000+1-1\/2001...
