1\/1乘2乘3+1\/2乘3乘4+...+1\/98乘99乘100=
1\/(n-1)n-1\/n(n+1)=(n+1-n+1)\/(n-1)n(n+1)=2\/(n-1)n(n+1),于是1\/(n-1)n(n+1)=(1\/2)[1\/(n-1)n-1\/n(n+1)]所以原式=(1\/2)[(1\/1x2-1\/2x3)+(1\/2x3-1\/3x4)+++++(1\/97x98-1\/98x99)+(1\/98x99-1\/99x100)]=(1\/2)(1\/2-1\/9900)...
小学奥数题1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(99*100)
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+1\/(5*6)+……+1\/(98*99)+1\/(99*100)=1-1\/2+1\/2-1\/3+...+1\/99-1\/100 =1-1\/100 =99\/100
奥数题1\/1×2×3+1\/2×3×4+……+1\/98×99×100
1\/(1×2×3)+1\/(2×3×4)+……+1\/(98×99×100)=1\/2*[(1\/(1*2)-1\/(2*3)+1\/(2*3)-1\/(3*4)+……+1\/(98*99-1\/(99*100)]=1\/2*[1\/2-1\/9900]=1\/4-1\/19800=(4950-1)\/19800=4949\/19800
小学奥数题:1*2+2*3+3*4+……99*100=?
99*100=(1\/3)*[99*100*101-98*99*100]1*2+2*3+3*4+……99*100 =(1\/3)*[99*100*101]=333300 =1*(1+1)+2*(2+1)+3*(3+1)+...99*(99+1)=1的平方+2的平方+3的平方+...99方+1+2+3+...99 =1\/6*99*(99+1)*(2*99+1)+(1+99)99\/2 =333300 ...
六年级数学奥数题1+1*2\/1+3*4\/1...99*100\/1
1+1\/1*2+1\/2*3+1\/3*4+……+1\/99*100 =1+1-1\/2+1\/2-1\/3+1\/3-1\/4+……+1\/99-1\/100 =2-1\/100 =1.99
小学数学简便计算奥数题1×1\/2+2×1\/3+3×1\/4+4×1\/5……99×1\/100
=(1-1\/2)X1+(1\/2-1\/6)X2+(1\/3-1\/12)X3+(1\/4-1\/20)X4...(1\/99-1\/9900)X99 =1-1\/2X1+ 1-1\/6X2+ 1-1\/12X3+ 1-1\/20X4... 1-1\/9900X99 =1-1\/2 + 1-1\/3 + 1-1\/4+ 1-1\/5... +1-1\/100 =1+1+1+1...1 - (1\/2+1\/3+1\/4+1\/5+...1\/100...
小学奥数题1\/3 1\/15 1\/35 1\/63 1\/99?
思路:观察1\/15=2×(1\/3-1\/5)1\/35=2×(1\/5-1\/7)……分解之后通过各个数相加可以抵消掉一部分,剩下的就很容易计算了。
求教奥数题:分子是1+1\/2*3+1\/3*5+1\/4*7+...+1\/50*99,分母是1\/51+1\/52...
分母式 1\/51 +1\/52 +1\/53 +1\/54 +……+1\/100 就相对简明,这50个分母就是 51到100 的50个连续的自然数。分子式 1+ 1\/2*3 +1\/3*5 +1\/4*7 +……+ 1\/50*99,这50个分数的分母,依次是 1、2、3、4、……50 这50个自然数,分别乘以 1、3、5、7、……99 这50个奇数。
数学奥数题:1\/1*2+1\/2*3+1\/3*4...+1\/2011*2012+1\/2012*2013
1\/1*2+1\/2*3+1\/3*4...+1\/2011*2012+1\/2012*2013 =1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/2011-1\/2012+1\/2012-1\/2013 =1-1\/2013 =2012\/2013
数学题目:1\/2+1\/3+1\/4+1\/5...1\/98+1\/99+1\/100=?高手来解决啊!
这题很简单呀!f(x)的通项是1\/x,从x=2到x=100 用积分就可以了,那个积分号我不知道怎么打出来,就用S代替吧 原式=S(从2到100)1\/x=lnx(从2 到100)=ln100-ln2=2ln10-ln2 完毕