题目内容
计算:
+
+
+
+
+
+
+
+
+…+
+
+…+
+
+
+…+
+
.
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 1999 |
| 2 |
| 1999 |
| 1998 |
| 1999 |
| 1999 |
| 1999 |
| 1998 |
| 1999 |
| 2 |
| 1999 |
| 1 |
| 1999 |
因为
+
+
+…+
+
+
+…
=
=
=k,
所以:
+
+
+
+
+
+
+
+
+…+
+
+…
+
+…+
+
,
=1+2+3+4+…+1999,
=(1+1999)×[(1999-1)÷1+1]÷2,
=1999×2000÷2,
=1999000.
| 1 |
| k |
| 2 |
| k |
| 3 |
| k |
| k |
| k |
| k-1 |
| k |
| k-2 |
| k |
| 1 |
| k |
| 1+2+3+…+k+k-1+k-2+k-3+…+1 |
| k |
| k(k+1)-k |
| k |
所以:
| 1 |
| 1 |
| 1 |
| 2 |
| 2 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 2 |
| 3 |
| 3 |
| 3 |
| 2 |
| 3 |
| 1 |
| 3 |
| 1 |
| 1999 |
| 2 |
| 1992 |
| 1998 |
| 1999 |
| 1999 |
| 1999 |
| 2 |
| 1999 |
| 1 |
| 1999 |
=1+2+3+4+…+1999,
=(1+1999)×[(1999-1)÷1+1]÷2,
=1999×2000÷2,
=1999000.
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