题目内容
计算
(1)1+2+4+8+16+32+64+128+256;
(2)1+
+
+
+
+
+
+
+
.
(1)1+2+4+8+16+32+64+128+256;
(2)1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
考点:加减法中的巧算
专题:计算问题(巧算速算)
分析:(1)后一个数是前一个数的2倍,原式乘上2,再减去原式,然后再计算;
(2)前一个数是后一个数的2倍,原式乘上2,再减去原式,然后再进一步计算.
(2)前一个数是后一个数的2倍,原式乘上2,再减去原式,然后再进一步计算.
解答:
解:(1)1+2+4+8+16+32+64+128+256
=2×(1+2+4+8+16+32+64+128+256)-(1+2+4+8+16+32+64+128+256)
=(2+4+8+16+32+64+128+256+512)-(1+2+4+8+16+32+64+128+256)
=(2+4+8+16+32+64+128+256)-(2+4+8+16+32+64+128+256)+512-1
=512-1
=511;
(2)1+
+
+
+
+
+
+
+
=2×(1+
+
+
+
+
+
+
+
)-(1+
+
+
+
+
+
+
+
)
=(2+1+
+
+
+
+
+
+
)-(1+
+
+
+
+
+
+
+
)
=(1+
+
+
+
+
+
+
)-(1+
+
+
+
+
+
+
)+2-
=2-
=1
.
=2×(1+2+4+8+16+32+64+128+256)-(1+2+4+8+16+32+64+128+256)
=(2+4+8+16+32+64+128+256+512)-(1+2+4+8+16+32+64+128+256)
=(2+4+8+16+32+64+128+256)-(2+4+8+16+32+64+128+256)+512-1
=512-1
=511;
(2)1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=2×(1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=(2+1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=(1+
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 8 |
| 1 |
| 16 |
| 1 |
| 32 |
| 1 |
| 64 |
| 1 |
| 128 |
| 1 |
| 256 |
=2-
| 1 |
| 256 |
=1
| 255 |
| 256 |
点评:认真观察,根据数字特点进行组合,从而达到巧算的目的.
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