我知道答案,只要过程。
计算:1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+(1\/5+2\/5+3\/5+4\/5)+…+(1\/60+2\/...
1\/3+2\/3=1 1\/4+2\/4+3\/4=2\/4+1=1\/2+1 1\/5+2\/5+3\/5+4\/5=(1\/5+4\/5)+(2\/5+3\/5)=1+1 1\/6+2\/6+3\/6+4\/6+5\/6=(1\/6+5\/6)+(2\/6+4\/6)+3\/6=1+1+1\/2 应该能看出规律了吧 那么7为分母的结果为1+1+1 8为分母的结果为1+1+1+1\/2 9的为...
1\/2+2\/3+3\/4+4\/5+5\/6+6\/7+7\/8=?
1479\/280
1\/2+2\/2*3+3\/2*3*4+4\/2*3*4*5+5\/2*3*4*5*6+6\/2*3*4*5*6*7=?
1\/2+2\/2*3+3\/2*3*4+4\/2*3*4*5+5\/2*3*4*5*6+6\/2*3*4*5*6*7 =(2-1)\/2+(3-1)\/2*3+(4-1)2*3*4+(5-1)2*3*4*5+(6-1)2*3*4*5*6+(7-1)2*3*4*5*6*7 =2\/2-1\/2+3\/2*3-1\/2*3+4\/2*3*4-1\/2*3*4+5\/2*3*4*5-1\/2*3*4*5+6\/2*...
1÷2+2÷3+3÷4+4÷5+5×6+6÷7+7÷8+8÷9+9÷10……+2022÷2023
根据Newton(牛顿)的幂级数有:ln(1+1\/x) = 1\/x - 1\/2x² + 1\/3x³ - ...于是:1\/x = ln((x+1)\/x) + 1\/2x² - 1\/3x³ + ...代入x=1,2,...,n,就给出:1\/1 = ln(2) + 1\/2 - 1\/3 + 1\/4 -1\/5 + ...1\/2 = ln(3\/2) + ...
1+1\/2+1\/3+2\/3+1\/4+2\/4+3\/4+1\/5+2\/5+3\/5+4\/5+...+1\/100+2\/100+...+...
从第2个数开始,数的分子可以看成是a1=1,d=1的等差数列的n项和,也等于(n+1)n\/2 而分母则看成n+1,相比则等于n\/2 1\/2+(1+2)\/3++(1+2+3)\/4+(1+2+3+4)\/5+...+(1+2+3...+99)\/100 =1\/2+2\/2+3\/2+4\/2+...+99\/2 =(1+2+3...+99)\/2 =(99+...
(1+2\/3+2+3\/4+3+4\/5+4+5\/6+5+6\/7+6+7\/8)\/(3+1\/3+5+2\/4+7+3\/5+9+4\/...
你把他全部换成假分数 上面是5\/3,11\/4,19\/5,29\/6,41\/7,55\/8 下面是10\/3,22\/4,38\/5,58\/6,82\/7,110\/8 下面是上面的2倍 所以是0.5
1+1\/2+2\/3+3\/4+4\/5+5\/6+6\/7+7\/8+8\/9
!每一个加数的分母都是下一个加数的分子!!而且若把加数按下面编号:a1=1 a2=1\/2 a3=2\/3 ..a9=8\/9 那么an=(n-1)\/n=1-1\/n 若想求和:1+1\/2+2\/3+3\/4+4\/5+5\/6+6\/7+7\/8+8\/9 =1+1-1\/2+1-1\/3+1-1\/4..+1-1\/9 =9-(1\/2+1\/3+1\/4+..+1\/9)...
1*2+2*3+3*4+4*5+5*6+6*7+...n*(n+1)如何求和?
解:令数列an=n*(n+1),那么1*2+2*3+3*4+4*5+5*6+6*7+...n*(n+1)即为数列an前n项和Sn。又因为an=n*(n+1)=n^2+n,那么Sn=1*2+2*3+3*4+4*5+5*6+6*7+...n*(n+1)=1^2+1+2^2+2+3^2+3+...+(n-1)^2+(n-1)+n^2+n =(1^2+2^2+3^2+......
1\/2+2\/2*3+3\/2*3*4+4\/2*3*4*5+5\/2*3*4*5*6如何简算
……5\/2*3*4*5*6=(6-1)\/(2*3*4*5*6)=1\/(2*3*4*5)-1\/(2*3*4*5*6)所以 1\/2+2\/2*3+3\/2*3*4+4\/2*3*4*5+5\/2*3*4*5*6 =1\/2+1\/2-1\/(2*3)+1\/(2*3)-1\/(2*3*4)+……+1\/(2*3*4*5)-1\/(2*3*4*5*6)=1-1\/(2*3*4*5*6)=719\/720 ...
1\/2+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+(1\/5+2\/5+3\/5+4\/5)+...
…+99\/100 =(1\/2)+(1\/3+2\/3)+(1\/4+2\/4+3\/4)+(1\/5+2\/5+3\/5+4\/5)+……+(1\/100+2\/100+……+99\/100)=1\/2+2\/2+3\/2+4\/2+...+99\/2 =[(1+99)*99\/2]\/2 =9900\/4 =2475 同学你好,如果问题已解决,记得右上角采纳哦~~~您的采纳是对我的肯定~谢谢哦 ...