(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/11*13)具体计算过程
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![(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/11*13)具体计算过程](/uploads/image/z/1987534-46-4.jpg?t=%281%2B1%2F1%2A3%29%2A%281%2B1%2F2%2A4%29%2A%281%2B1%2F3%2A5%29%2A.%2A%281%2B1%2F11%2A13%29%E5%85%B7%E4%BD%93%E8%AE%A1%E7%AE%97%E8%BF%87%E7%A8%8B)
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/11*13)具体计算过程
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/11*13)具体计算过程
(1+1/1*3)*(1+1/2*4)*(1+1/3*5)*.*(1+1/11*13)具体计算过程
可以先假定任意一项为 1+1/[n(n+2)],可以将该项变换为 1+1/[n(n+2)] = [n(n+2)+1]/[n(n+2)] = (n^2=2n+1)/[n(n+2)] = (n+1)^2/[n(n+2)];这样,就可以得到:
原式 = 2^2/(1x3)x3^2/(2x4)x4^2/(3x5)x...x12^2/(11x13) = 2/1x12/13 = 24/13.
中间的分子分母抵消过程,自己写出来就清楚了.