求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2
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![求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2](/uploads/image/z/1762971-51-1.jpg?t=%E6%B1%82%5B%EF%BC%881%2A2%2A4%2B2%2A4%2A8%2B%E2%80%A6%2Bn%2A2n%2A4n%EF%BC%89%2F%281%2A3%2A9%2B2%2A6%2A18%2B...%2Bn%2A3n%2A9n%29%5D%5E2)
求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2
求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2
求[(1*2*4+2*4*8+…+n*2n*4n)/(1*3*9+2*6*18+...+n*3n*9n)]^2
提公因式1*2*4和1*3*9得:
1*2*4(1^3+2^3+...n^3)/[1*3*9(1^3+2^3+...n^3)]^2
即:1*2*4(1^3+2^3+...n^3)/(1*3*9)^2*(1^3+2^3+...n^3)^2
=8(1^3+2^3+...n^3)/27^2*(1^3+2^3+...n^3)^2
上下消去(1^3+2^3+...n^3)得:8/729(1^3+2^3+...n^3)
又(1^3+2^3+...n^3)=[n(n+1)/2]^2,得:
原式=32/729[n(n+1)]^2
不好意思,算错了。不算了。