37 lines
1.1 KiB
Markdown
37 lines
1.1 KiB
Markdown
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---
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title: Sum square difference
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localeTitle: 和方差
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---
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## 问题6：求和方差
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### 方法：
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* 前n个自然数的总和可以使用以下公式计算：
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* 
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* n个自然数的平方和可以使用以下公式计算：
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* 
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* 我们可以使用上面的公式计算值并减去它们以得到结果。
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### 解：
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```js
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function sumSquareDifference(n) {
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const sumOfN = (n*(n+1))/2;
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const sumOfNSquare = (n*(n+1)*(2*n+1))/6;
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//** is exponentaial operator added in ES7, it's equivalent to Math.pow(num, 2)`
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return (sumOfN ** 2) - sumOfNSquare;
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}
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```
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* [运行代码](https://repl.it/@ezioda004/Problem-6-Sum-square-difference)
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### 参考文献：
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* [n个数字的总和 - 维基百科](https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF)
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* [n平方数的总和 - 维基百科](https://en.wikipedia.org/wiki/Square_pyramidal_number)
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