1十3十5十7十…十17十19脱式用简便算法? 1十3十5十7十…十17十19脱式用简便算法
1\u53413\u53415\u53417\u5341\u2026\u534117\u534119\u8131\u5f0f\u7528\u7b80\u4fbf\u7b97\u6cd5\uff1f\u65b9\u6cd5\u4e00\uff1a1+3+5+7+9+11+13+15+17+19\u7684\u7b80\u4fbf\u8fd0\u7b97
=\uff081+19\uff09\u00d75
=20\u00d75
=100
\u65b9\u6cd5\u4e8c\uff1a
1+3+5+7+9+11+13+15+17+19
=\uff081+19\uff09\u00d710\u00f72
=20\u00d710\u00f72
=200\u00f72
=100
1\u53413\u53415\u53417\u5341\u2026\u534117\u534119\u8131\u5f0f\u7528\u7b80\u4fbf\u7b97\u6cd5
1\u53413\u53415\u53417\u5341\u2026\u534117\u534119
=\uff081+19\uff09+\uff083+17\uff09+\uff085+15\uff09+...
=20*5
=100
=(1+19)×5
=20×5
=100
您好!为您解答如下
1.观察这些数字,第一个加最后一个数字是20. 第二个数字加倒数第二个数字也是20。以后每2个数字之和都是20。简便算法是 :20乘以项数除以2
2.计算:
1+3+5+7+ ……17+19
=(1+19)* 10 /2
= 100
因为公差是2,收尾两个数相加都是1+19=20,一共有10个数,所以和就是10×20÷2=100
1+3+5+7+9+……+17+19
=(1+19)×(10÷2)
=20×5
=100
绛旓細绠渚胯繍绠楀氨鏄噾鏁达紝閭f垜浠湅涓綅鏁扮浉鍔犱负10鐨勫厛杩涜璁$畻姣旇緝鏂逛究鍙g畻锛岃繍鐢ㄥ姞娉曚氦鎹㈠緥锛屽彉鎹綅缃笉褰卞搷璁$畻缁撴灉 锛3+7锛+锛5+15锛+锛11+19锛+17 =10+20+30+17 =77 璇烽噰绾宠阿璋
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