Question

1/(1×3×5) +1/(3×5×7)+1/(5×7×9)+…+1/(11×13×15)这个式子怎么用简便运算

Answer 1

把通项1/(a×b×c)改写成(1/a-2/b+1/c)/8,然后中间的项可以加减抵消掉，只需计算两端的若干个分数的运算即可。
1/(1×3×5) +1/(3×5×7)+…+1/(11×13×15)
=(1/1-2/3+1/5)/8+(1/3-2/5+1/7)/8+...+(1/11-2/13+1/15)/8
=(1/1-2/3+1/5+1/3-2/5+1/7+...+1/11-2/13+1/15)/8
=(2/3-1/13+1/15)/8
=16/195
不知大家有无更好的方法。

Answer 2

上面这位兄弟解法是对的，就是倍数弄错了。。。。

1/(1×3×5) +1/(3×5×7)+1/(5×7×9)+…+1/(11×13×15)

=(1/4)×[1/(1×3)-1/(3×5)]+(1/4)×[1/(3×5)-1/(5×7)]+....+(1/4)×[1/(11×13)-1/(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)]

=(1/4)×(64/195)

=16/195

Answer 3

1/(1×3×5) +1/(3×5×7)+1/(5×7×9)+…+1/(11×13×15)

=1/2x[1/1x3-1/3x5]+1/2[1/3x5-1/5x7]+....+1/2[1/11x13-1/13x15]

=1/2[1/1x3-1/3x5+1/3x5-1/5x7+....+1/11x13-1/13x15]

=1/2x[1/1x3-1/13x15]

=1/2x64/195

=32/195