作业帮[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/02 17:00:48
![[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?](/uploads/image/z/11547341-53-1.jpg?t=%5B%282%5E3-1%29%283%5E3-1%29...100%5E3-1%29%5D%2F%5B%282%5E3%2B1%29%283%5E3%2B1%29...%28100%5E3%2B1%29%5D%3D%3F)
[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?
[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?
[(2^3-1)(3^3-1)...100^3-1)]/[(2^3+1)(3^3+1)...(100^3+1)]=?
100很大,你可以设通项为(n^3-1)/(n^3+1),原式为n从2到100通项的乘积 ∏(n^3-1)/(n^3+1) =∏(n-1)[n(n+1)+1]/{(n+1)[(n-1)n+1]} =∏[(n-1)/(n+1)]*∏{[n(n+1)+1]/[(n-1)n+1]} ={2/[n(n+1)]}{[n(n+1)+1]/(1*2+1)} =(2/3){[n(n+1)+1]/[n(n+1)]} [n(n+1)+1]/[n(n+1)]当n很大时,趋于1 ∴(2/3){[n(n+1)+1]/[n(n+1)]}趋于2/3
10、(1)(2)(3)
2/3×3/4×10×1/2
化简 1*1!+2*2!+3*3!+...+10*10!
1+2+9+5+3-10+3
1+2+3+4*10
1+2+3+.+10=
9,10,1,2,3
1+1/2+2/2+1/2+1/3+2/3+3/3+2/3+1/3+.+1/10+2/10+3/10+.+9/10+10/10+9/10+...+2/10+1/10=?
1+1+2+1+2+3+1+2+3+4+.+1+2+3+...+10急
(1-1/2*2)*(1-1/3*3)*…...*(1-1/10*10)=?
求和2(1-1/2)+3(1-1/3)+……+10(1-1/10)=
|1-1/2|-|1/2-1/3|-|1/3-1/4|-.-|1/9-1/10|
|1-1/2|+|1/2-1/3|+|1/3-1/4|+...|1/9-1/10|
计算1又1/2-1/3+2又1/2-2/3+3又1/2-3/3+……+10又1/2-10/3
1/1*2*3+1/2*3*4+1/3*4*5+.+1/8*9*10
(1+2X/3)-10-3X/2=1
计算(3*10^-5)^2/(3*10^-1)^2
[(1+3)(1+3^2)(1+3^4)(1+3^8).(1+3^32)+1/2]3*3^2*3^3*……*3^10=多少?错了,应为[(1+3)(1+3^2)(1+3^4)(1+3^8).........(1+3^32)+1/2]/3*3^2*3^3*……*3^10=多少,中括号后少了个除号