-1/3-1/7-1/9-1/11
=1+(6+3+2+1)/12+(5+4)/40
-(3+1)/9-(11+7)/77
=1+1+9/40-4/9-18/77
=9/40+5/9+59/77
=(9×9×77+5×40×77+59×40×5)/(360×77)
=33437/27720
=1+5717/27720
=1+7/8-4/9-1/10+1/4-(1/7+1/11)
=1+(7/8+1/4)-(4/9+1/10)-18/77
=1+9/8-49/90-18/77
=9/8-18/77+1-49/90
=549/616+41/90
=37333/27720
1+1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9+1\/10-1\/11+1\/12 等于多少 要分数的...
原式=1+1\/2+1\/4+1\/6+1\/8+1\/10+1\/12 -1\/3-1\/7-1\/9-1\/11 =1+(6+3+2+1)\/12+(5+4)\/40 -(3+1)\/9-(11+7)\/77 =1+1+9\/40-4\/9-18\/77 =9\/40+5\/9+59\/77 =(9×9×77+5×40×77+59×40×5)\/(360×77)=33437\/27720 =1+5717\/27720 ...
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10=多少
计算1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10的过程如下：首先，我们可以通过拆分求和的方法简化计算：= 1 + (1\/2 + 1\/3 + 1\/6) + (1\/4 + 1\/5 + 1\/10) + 1\/7 + 1\/8 + 1\/9 = 1 + 1 + (1\/2 + 1\/3 - 1\/6) + (1\/4 + 1\/5 + 1\/10)= 1 +...
...1\/4+1\/6-1\/5+1\/7-1\/6+1\/8-1\/7+1\/9-1\/8+1\/10-1\/9=?
原式 = 1 + 1\/3 + 1\/4 + 1\/5 + 1\/6 + 1\/7 + 1\/8 + 1\/9 + 1\/10 - 1\/2 - 1\/3 - 1\/4 - 1\/5 - 1\/6 - 1\/7 - 1\/8 - 1\/9 = 1 + 1\/10 - 1\/2 = 1\/2 + 1\/10 = 3\/5 希望对你有帮助，新年快乐~...
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10。计算结果。
=1+(1\/2+1\/3+1\/6)+(1\/4+1\/5+1\/10)+1\/7+1\/8+1\/9 =1+1+11\/20+1\/8+1\/7+1\/9 =1+1+27\/40+1\/9+1\/7 =1+1+283\/360+1\/7 =1+1+2341\/2520 =1+4861\/2520 ≈2.92896
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10的结果大于哪个整数而小于哪个...
主要过程...1\/2=0.5，1\/3约为0.34，1\/4=0.25，1\/5=0.2，1\/6约为0.17，1\/7+1\/8+1\/9+1\/10约为0.48（麻烦了，自己算吧）1+0.5+0.34+0.25+0.2+0.17+0.48=2.94 又因为取得近似值所以小于3大于2
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+...1\/n等于多少?
1\/x = ln((x+1)\/x) + 1\/2x^2 - 1\/3x^3 + ...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\/2*4 - 1\/3*8 + 1\/4*16 - ...1\/n = ln((n+1)\/n) + 1\/2n^2 - 1\/3n^3 + ...相加，就得到...
1+1\/2+1\/3+1\/4+1\/5+1\/6+...+1\/n极限多少?(过程)
1\/5+1\/6+1\/7+1\/8>1\/8+1\/8+1\/8+1\/8=1\/2 1\/9+1\/10+……+1\/16>1\/16+1\/16+……+1\/16=1\/2 ……所以1+1\/2+1\/3+1\/4+1\/5+1\/6+...+1\/n >1+1\/2+1\/2+1\/2+1\/2+……(无穷多个1\/2相加)所以1+1\/2+1\/2+1\/2+1\/2+……是无穷大 所以1+1\/2+1\/3+1\/...
...1+1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9+1\/10 在线等高手解答
符号变换 { item=mark*1.0\/i; \/\/每一项 sum+=item; \/\/加到一起 mark=-mark; \/\/每一项的正负号变换 } return sum;} int main(){ printf("1+1\/2-1\/3+1\/4-1\/5+1\/6-1\/7+1\/8-1\/9+1\/10=%lf\\n",func(10));} ...
1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10的循环节是多少?
然后进行运算。1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10等于(10+9+8+7+6+5+4+3+2+1)\/10。这道题的答案是循环节。循环节是指数列中出现的重复的数字。在这道题中，循环节是1\/10。因此，1+1\/2+1\/3+1\/4+1\/5+1\/6+1\/7+1\/8+1\/9+1\/10的循环节是1\/10。
1-1\/2+1\/3-1\/4+1\/5-1\/6+1\/7-1\/8+1\/9-1\/10+1\/11-1\/12
一就是淡定慢慢算，，可以求数列1\/n的和与1\/2n的和作差吧
