题目内容
计算:1×
+2×
+3×
+…27×
+28×
= .
| 2 |
| 29 |
| 3 |
| 29 |
| 4 |
| 29 |
| 28 |
| 29 |
| 29 |
| 29 |
考点:分数的巧算
专题:计算问题(巧算速算)
分析:先把原式化为=(1×2+2×3+3×4+…+27×28+28×29)×
,括号内的运算如下:
1×(1+1)+2×(1+2)+3×(1+3)+…28×(1+28),然后运用乘法分配律计算,变为:(1+2+3+…28)+(1×1+2×2+3×3+…28×28),再运用以下公式计算即可:1+2+3+…+n=(1+n)×n÷2,1×1+2×2+3×3+4×4+…+n×n=n(n+1)(2n+1)÷6.
| 1 |
| 29 |
1×(1+1)+2×(1+2)+3×(1+3)+…28×(1+28),然后运用乘法分配律计算,变为:(1+2+3+…28)+(1×1+2×2+3×3+…28×28),再运用以下公式计算即可:1+2+3+…+n=(1+n)×n÷2,1×1+2×2+3×3+4×4+…+n×n=n(n+1)(2n+1)÷6.
解答:
解:1×
+2×
+3×
+…27×
+28×
=(1×2×
)+(2×3×
+3×4×
)+…+(27×28×
+28×29×
)
=(1×2+2×3+3×4+…+27×28+28×29)×
=[1×(1+1)+2×(1+2)+3×(1+3)+…28×(1+28)]×
=[1+1×1+2+2×2+3+3×3+…28+28×28]×
=[(1+2+3+…28)+(1×1+2×2+3×3+…28×28)]×
=[(1+28)×28÷2+28×(28+1)×(2×28+1)÷6]×
=[406+7714]×
=8120×
=280
故答案为:280.
| 2 |
| 29 |
| 3 |
| 29 |
| 4 |
| 29 |
| 28 |
| 29 |
| 29 |
| 29 |
=(1×2×
| 1 |
| 29 |
| 1 |
| 29 |
| 1 |
| 29 |
| 1 |
| 29 |
| 1 |
| 29 |
=(1×2+2×3+3×4+…+27×28+28×29)×
| 1 |
| 29 |
=[1×(1+1)+2×(1+2)+3×(1+3)+…28×(1+28)]×
| 1 |
| 29 |
=[1+1×1+2+2×2+3+3×3+…28+28×28]×
| 1 |
| 29 |
=[(1+2+3+…28)+(1×1+2×2+3×3+…28×28)]×
| 1 |
| 29 |
=[(1+28)×28÷2+28×(28+1)×(2×28+1)÷6]×
| 1 |
| 29 |
=[406+7714]×
| 1 |
| 29 |
=8120×
| 1 |
| 29 |
=280
故答案为:280.
点评:完成此题,关键运用了一下两个公式:1+2+3+…+n=(1+n)×n÷2,1×1+2×2+3×3+4×4+…+n×n=n(n+1)(2n+1)÷6,
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