在大多数情况下,关键加法/减法技巧如下:据/p>
加法的顺序是不相关的,所以一个给定的和可以以任何想要的方式重新排列。据/p>
这种方法的通常应用是寻找加起来是10倍数的数字,因为这些数字很容易可视化和处理。据/p>
确定…的价值据/p>
3.据/span>+据/span>8.据/span>+据/span>6.据/span>+据/span>2据/span>+据/span>1据/span>7.据/span>+据/span>3.据/span>+据/span>5.据/span>+据/span>1据/span>+据/span>2据/span>+据/span>4.据/span>.据/span>
添加可以直接进行,没有太多麻烦,但这种过程相对繁琐,易受错误。更清洁的方法是重新排列术语:据/p>
3.据/span>+据/span>8.据/span>+据/span>6.据/span>+据/span>2据/span>+据/span>1据/span>7.据/span>+据/span>3.据/span>+据/span>5.据/span>+据/span>1据/span>+据/span>2据/span>+据/span>4.据/span>=据/span>3.据/span>+据/span>1据/span>7.据/span>+据/span>8.据/span>+据/span>2据/span>+据/span>6.据/span>+据/span>4.据/span>+据/span>3.据/span>+据/span>5.据/span>+据/span>2据/span>+据/span>1据/span>=据/span>2据/span>0.据/span>+据/span>1据/span>0.据/span>+据/span>1据/span>0.据/span>+据/span>1据/span>0.据/span>+据/span>1据/span>=据/span>5.据/span>1据/span>.据/span>□据/span>
拿1000,然后加20。据B.R.>现在再加1000。据B.R.>现在添加30。据B.R.>现在再加1000。据B.R.>现在加40。据B.R.>现在再加1000。据B.R.>现在添加10。据/p>
确定…的价值据/p>
1据/span>+据/span>9.据/span>+据/span>1据/span>8.据/span>+据/span>3.据/span>+据/span>5.据/span>+据/span>2据/span>+据/span>1据/span>7.据/span>+据/span>1据/span>5.据/span>+据/span>2据/span>+据/span>8.据/span>.据/span>
鉴于Number 100,您任务要执行以下操作:据/p>
- 首先,向数量添加7据/li>
- 其次,添加10到数字据/li>
- 第三,在数字上加上6据/li>
- 第四,向数量添加10据/li>
- 第五,向数量添加14据/li>
- 第六,在数字上加上10据/li>
- 最后,将13添加到数字据/li>
结果是什么?据/strong>
另一种技术在寻找以下某些模式之后的系列和据一种href="//www.parkandroid.com/wiki/arithmetic-progressions/" class="wiki_link" title="等差数列" target="_blank">等差数列据/a>或者据一种href="//www.parkandroid.com/wiki/geometric-progression-sum/" class="wiki_link" title="几何进步" target="_blank">几何进步据/a>.一般的技术是将和操作成另一个相关的和,并将这两个和一起使用来取消项和/或实现期望的重复。下面是一个例子:据/p>
高斯的老师让他把1到100之间的所有整数(包括100在内)加起来。这个和是多少?据/p>
关键的实现是重新排列术语,使数字配对:据/p>
S.据/span>=据/span>1据/span>+据/span>2据/span>+据/span>3.据/span>+据/span>⋯据/span>+据/span>9.据/span>8.据/span>+据/span>9.据/span>9.据/span>+据/span>1据/span>0.据/span>0.据/span>=据/span>(据/span>1据/span>+据/span>1据/span>0.据/span>0.据/span>)据/span>+据/span>(据/span>2据/span>+据/span>9.据/span>9.据/span>)据/span>+据/span>(据/span>3.据/span>+据/span>9.据/span>8.据/span>)据/span>+据/span>⋯据/span>+据/span>(据/span>5.据/span>0.据/span>+据/span>5.据/span>1据/span>)据/span>=据/span>1据/span>0.据/span>1据/span>×据/span>5.据/span>0.据/span>=据/span>5.据/span>0.据/span>5.据/span>0.据/span>.据/span>□据/span>
当有偶数术语时,这种技术运行良好。当存在奇数术语时,该技术基本相同,但具有稍微更聪明的实现:据/p>
现在高斯的老师让他把1到101之间的所有整数(包括1和101)加起来。这个和是多少?据/p>
“配对术语”的直觉可以被形式化:据/p>
S.据/span>=据/span>=据/span>2据/span>S.据/span>=据/span>=据/span>⇒据/span>S.据/span>=据/span>=据/span>1据/span>+据/span>2据/span>+据/span>3.据/span>+据/span>⋯据/span>+据/span>9.据/span>9.据/span>+据/span>1据/span>0.据/span>0.据/span>+据/span>1据/span>0.据/span>1据/span>1据/span>0.据/span>1据/span>+据/span>1据/span>0.据/span>0.据/span>+据/span>9.据/span>9.据/span>+据/span>⋯据/span>+据/span>3.据/span>+据/span>2据/span>+据/span>1据/span>(据/span>1据/span>+据/span>1据/span>0.据/span>1据/span>)据/span>+据/span>(据/span>2据/span>+据/span>1据/span>0.据/span>0.据/span>)据/span>+据/span>⋯据/span>+据/span>(据/span>5.据/span>0.据/span>+据/span>5.据/span>2据/span>)据/span>+据/span>(据/span>5.据/span>1据/span>+据/span>5.据/span>1据/span>)据/span>+据/span>(据/span>5.据/span>2据/span>+据/span>5.据/span>0.据/span>)据/span>⋯据/span>+据/span>(据/span>1据/span>0.据/span>0.据/span>+据/span>2据/span>)据/span>+据/span>(据/span>1据/span>0.据/span>1据/span>+据/span>1据/span>)据/span>1据/span>0.据/span>2据/span>×据/span>1据/span>0.据/span>1据/span>5.据/span>1据/span>⋅据/span>1据/span>0.据/span>1据/span>5.据/span>1据/span>5.据/span>1据/span>.据/span>□据/span>
这导致了一般结果:据/p>
一个据span class="katex">
N据/span>- 具有第一项的算术系列据span class="katex">
一种据/span>和上学期据span class="katex">
B.据/span>已经和据/p>
2据/span>N据/span>(据/span>一种据/span>+据/span>B.据/span>)据/span>.据/span>□据/span>
是什么据span class="katex">
N据/span>=据/span>0.据/span>∑据/span>2据/span>0.据/span>0.据/span>0.据/span>N据/span>还是据/span>
几何级数可以用类似的方法处理。下面是一个例子:据/p>
高斯的老师现在要求他求出2的前10次非负幂的和。这个和是多少?据/p>
类似于之前,计算据/p>
S.据/span>2据/span>S.据/span>⇒据/span>S.据/span>=据/span>2据/span>0.据/span>+据/span>2据/span>1据/span>+据/span>⋯据/span>+据/span>2据/span>9.据/span>=据/span>2据/span>1据/span>+据/span>2据/span>2据/span>+据/span>⋯据/span>+据/span>2据/span>1据/span>0.据/span>=据/span>2据/span>S.据/span>−据/span>S.据/span>=据/span>2据/span>1据/span>0.据/span>−据/span>2据/span>0.据/span>=据/span>1据/span>0.据/span>2据/span>3.据/span>那据/span>
其中一般策略要乘以常规比例,然后减去。据span class="katex">
□据/span>
这导致了一般结果:据/p>
一个据span class="katex">
N据/span>第一个术语 - 员几何系列据span class="katex">
一种据/span>和常见的比据span class="katex">
R.据/span>已经和据/p>
一种据/span>⋅据/span>R.据/span>−据/span>1据/span>R.据/span>N据/span>−据/span>1据/span>.据/span>□据/span>
1据/span>
2据/span>
3.据/span>
4.据/span>
考虑具有常见比率的几何进展据span class="katex">
4.据/span>.据/span>如果第一个的和据span class="katex">
5.据/span>术语是据span class="katex">
1据/span>0.据/span>2据/span>3.据/span>那据/span>什么是初期的术语?据/p>