1/1x3+1/3x5+1/5x7+…+1/(2n-1)(2n+1)的值为17/35,求n的值。
第二题有简单些的过程么?初中
追答楼下给了详细的解答。
1/n(n+1)=1/n-1/(n+1)
1/(2n-1)(2n+1)=[1/(2n-1)-1/(2n+1)]/2
1\/1x2+1\/2x3+1\/3x4+…+1\/n(n+1)=___,(用含有n的式子表示)
(1) n\/(n+1)(2) n=17 都是裂项求和
1\/1*2+1\/2*3+1\/3*4+...1\/n(n+1)
1\/1×2+1\/2×3+1\/3×4+...1\/n(n+1)==n\/n+1。1、可以分析数列的规律:1\/1×2=1-1\/2,1\/2×3=1\/2-1\/3;即每个数字都可以进行拆分为两个分数相减,通项公式为:1\/n(n+1)=1\/n-1\/n+1 2、1\/1×2+1\/2×3+1\/3×4+...1\/n(n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/...
1\/1x2+1\/2x3+1\/3x4+...+1\/n(n+1)=?
由公式 1\/nx(n+m)=m(1\/n-1\/n+m)所以1、1\/1-1\/2+1\/2-1\/3+1\/3-1\/4...1\/n-1\/n+1=1-1\/n+1=n\/n+1 (中间的每2项可以消去)2、2(1\/2-1\/4+1\/4-1\/6...1\/2010-1\/2012)=2*(1\/2-1\/2012)=1-1\/1006=1005\/1006 ...
1\/1×2+1\/2×3+1\/3×4+1\/4×5+…+1\/n(n+1)=?
解:原式=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+…+1\/n-1\/[n+1]=1-1\/[1+n]=n\/[n+1]分析:1\/1*2=1-1\/2 1\/2*3=1\/2-1\/3 从而得解
1\/1x2+1\/2x3+1\/3x4+...+1\/nx(n+1)=__
裂项相消法.1\/(1*2)+1\/(2*3)+1\/(3*4)+.+1\/[n(n+1)]=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+.+[1\/n-1\/(n+1)]=1-1\/(n+1)=n\/(n+1)填:n\/(n+1)
1\/1x2+1\/2x3+1\/3x4+...+1\/n(n+1)怎么算
1\/1x2+1\/2x3+1\/3x4+...+1\/n(n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)
1\/1*2+1\/2*3+1\/3*4+…+1\/n(n+1)
解:依题意得A1=1\/1*2,A2=1\/2*3...所以 An=1\/n(n+1)An=1\/n(n+1)=1\/n-1\/(n+1)所以A1=1-1\/(1+1)1-1\/2 A2=1\/2-1\/3 A3=1\/3-1\/4 。。。An=1\/n-1\/(n+1)所以1\/1*2+1\/2*3+1\/3*4+…+1\/n(n+1)就等于 A1+A2+A3+...+An= 1-1\/2+1\/2-1\/3+1...
一道数学题,请用简便方法计算:1\/1x2+1\/2x3+1\/3x4+...+1\/10x11_百度知 ...
10\/11。因为这个式子的每一项都是1\/n(n+1)的形式。且1\/n(n+1)=[(n+1)-n]\/n(n+1)=1\/n-1\/(n+1)所以原式=(1-1\/2)+(1\/2-1\/3)+……+(1\/10-1\/11)=1+(-1\/2+1\/2-1\/3+……+1\/10)-1\/11,括号内都相互抵消 =1-1\/11 =10\/11 乘除法 1、分数乘整数,分母不变...
求和:1\/1x2+1\/2x3+1\/3x4+···+1\/n(n+1)
1\/1x2+1\/2x3+1\/3x4+···+1\/n(n+1)=1\/2+1\/6+1\/12+...+1\/n(n+1)=1-1\/2+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/n-1\/n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/n-1\/n+1 =1-1\/n+1 =n\/n+1
1\/1x2+1\/2x3+1\/3x4+…+1\/2010x2011 讲解一下! 知道答案但看不明白啊...
1\/(1×2)=1-1\/2 1\/(2×3)=1\/2-1\/3 1\/1x2+1\/2x3+1\/3x4+…+1\/2010x2011 =1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/2010-1\/2011 =1+(1\/2-1\/2)+(1\/3-1\/3)+...+(1\/2010-1\/2010)-1\/2011 =1-1\/2011 =2010\/2011 裂项求和 ...