(1+0.1)+(2+0.1*2)+(3+0.1*3)+...+(99+0.1*99)+(100+0.1*100)
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(1+0.1)+(2+0.1*2)+(3+0.1*3)+...+(99+0.1*99)+(100+0.1*100)
![(1+0.1)+(2+0.1*2)+(3+0.1*3)+...+(99+0.1*99)+(100+0.1*100)](/uploads/image/z/16024901-5-1.jpg?t=%281%2B0.1%29%2B%282%2B0.1%2A2%29%2B%283%2B0.1%2A3%29%2B...%2B%2899%2B0.1%2A99%29%2B%28100%2B0.1%2A100%29)
(1+0.1)+(2+0.1*2)+(3+0.1*3)+...+(99+0.1*99)+(100+0.1*100)
=(1+2+3+……+99+100)+0.1*(1+2+3+……+99+100)
=(1+100)*100/2+0.1*(1+100)*100/2
=5050+505
=5555
=(1+2+3+……+99+100)+0.1*(1+2+3+……+99+100)
=(1+100)*100/2+0.1*(1+100)*100/2
=5050+505
=5555
(1+0.1)+(2+0.1*2)+(3+0.1*3)+...+(99+0.1*99)+(100+0.1*100)等于几
证明:1/2*3/4*5/6*7/8*.*97/98*99/100大于0.1
100分之99和0.1..3又2分之1和3.6..0.45和20分之9..比较大小
(2-0.1*3)+(4-0.1*5)+(6-0.1*7)+(8-0.1*9)+(10-0.1*11)+(12-0.1*
C语言程序设计,1、编写程序求1-1/2+1/3-1/4+.+1/99-1/100的值.2、一张纸厚度为0.1mm,珠穆
0.1*99等于
5.66×99+56.6×0.1
分子是1乘3乘5乘到99,分母是2乘4乘6乘到100,求这个数与0.1的大小关系,
1:49.9²+9.98+0.1² 2:99²+199
100个0.1相除:0.1除以0.1除以0.1除以0.1除以0.1……
0.8除以0.1乘0.1除以0.1乘0.1除以“\”“\”“【100个0.1】乘0.1
计算:1-0.1分之1-(0.1分之1)的2次幂-(0.1分之1)的3次幂