简便计算1/1*3*5+1/3*5*7+1/5*7*9+1/7*9*11+1/9*11*13+1/11*13*15中1/

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分析:1/1×3×5=1/4×(1/1×3 -1/3×5)
1/3×5×7=1/4×(1/3×5 -1/5×7)
1/5×7×9=1/4×(1/5×7 - 1/7×9)
1/7×9×11=1/4×(1/7×9 -1/9×11)
.
1/11×13×15=1/4×(1/11×13 -1/13×15)
所有的等式相加有
1/1×3×5+1/3×5×7+1/5×7×9+.+1/11×13×15
=1/4×(1/1×3 -1/3×5 +1/3×5 -1/5×7+.+1/11×13 -1/13×15)
=1/4×(1/1×3 - 1/13×15)
=16/195
结论:1/n(n+1)(n+2)=1/2×[1/n(n+1) - 1/(n+1)(n+2)]
1/n(n+2)(n+4)=1/4×[1/n(n+2) - 1/(n+2)(n+4)]

1/1*3*5+1/3*5*7+1/5*7*9+1/7*9*11+1/9*11*13+1/11*13*15
=1/2*（1/1*3-1/3*5）+1/2*（1/3*5-1/5*7）+1/2*（1/5*7-1/7*9）+1/2*（1/7*9-1/9*11）+1/2*（1/9*11-1/11*13）+1/2*（1/11*13-1/13*15）
=1/2*（1/1*3-1/3*5+1/3...