题目内容
(1+
)+(2+
)+(3+
)+…+(97+
)．
| 1 |
| 97 |
| 2 |
| 97 |
| 3 |
| 97 |
| 97 |
| 97 |
分析：(1+
)+(2+
)+(3+
)+…+(97+
)还可以写成(1+
)+2×(1+
)+3×(1+
)+…97×(1+
)，然后利用分配律变式为（1+2+3…+97）×(1+
)的形式，计算即可解答．
| 1 |
| 97 |
| 2 |
| 97 |
| 3 |
| 97 |
| 97 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
解答：解：(1+
)+(2+
)+(3+
)+…+(97+
)
=(1+
)+2×(1+
)+3×(1+
)+…97×(1+
)，
=（1+2+3…+97）×(1+
)，
=
×(1+
)
=
×1+
×
=4753+49
=4802
| 1 |
| 97 |
| 2 |
| 97 |
| 3 |
| 97 |
| 97 |
| 97 |
=(1+
| 1 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
| 1 |
| 97 |
=（1+2+3…+97）×(1+
| 1 |
| 97 |
=
| 97×(97+1) |
| 2 |
| 1 |
| 97 |
=
| 97×(97+1) |
| 2 |
| 97×(97+1) |
| 2 |
| 1 |
| 97 |
=4753+49
=4802
点评：本题主要考查分数的巧算，熟练利用运算律是解答本题的关键．
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