正整数平方及更高幂次公式简单推导

首先是$\sum\limits_{i=1}^n i$

因为$(n+1)^2-n^2=2\times n+1$

$\begin{align} (n+1)^2-n^2+n^2-(n-1)^2+\cdots +2^2-1^2 & = 2n+1+2(n-1)+1+\cdots+2+1 \end{align}$

$\begin{align} (n+1)^2-1=2\times \sum_{i=1}^n i+n \ \end{align}$

$\begin{align} \frac{n(n+1)}{2}=\sum\limits_{i=1}^n i\ \end{align}$

之后推导到平方上去。

$(n+1)^3-n^3=3n^2+3n+1$

$\to (n+1)^3-n^3+n^3-(n-1)^3+\cdots +2^3-1^3=3\sum\limits_{i=1}^ni^2+3\sum\limits_{i=1}^ni+n$

$\to (n+1)^3-1-\frac{3n(n+1)}{2}-n=3\sum\limits_{i=1}^n i^2$

$\to (n+1)[(n+1)^2-\frac{3n}{2}-1]=3\sum\limits_{i=1}^n i^2$

$\to \frac{(n+1)(2n+1)n}{6}=\sum\limits_{i=1}^n i^2$

以此类推

$\sum\limits_{i=1}^n i^k=$最高次数为$k+1$的$n$的多项式

以及附带独特版本：

已知离散求和与连续求积分之间的关系 (可以画图直观理解):

$\sum_{i=1}^n i^2 = \int_0^n x^2 \mathrm{d} x + \sum_{i=1}^n \int_{i-1}^i (i^2 -x^2) \mathrm{d} x$

因为：

$\begin{aligned}\int_{i-1}^i (i^2 -x^2) \mathrm{d} x &= \left. \left(i^2 x - \frac{1}{3}x^3\right) \right|_{i-1}^i \\ &= \left(i^3 - \frac{1}{3}i^3\right) - \left(i^2 (i-1) - \frac{1}{3}(i-1)^3 \right) \\ &= \frac{1}{3}(i-1)^3 -\frac{1}{3}i^3 + i^2 \\ &= \frac{1}{3} (i^3 - 3i^2 + 3i - 1) -\frac{1}{3}i^3 + i^2 \\ &= i - \frac{1}{3} \end{aligned}$

所以：

$\begin{aligned} \sum_{i=1}^n i^2 &= \int_0^n x^2 \mathrm{d} x + \sum_{i=1}^n \int_{i-1}^i (i^2 -x^2) \mathrm{d} x \\ &= \int_0^n x^2 \mathrm{d} x + \sum_{i=1}^n \left(i - \frac{1}{3}\right) \end{aligned}$

又因为：

$\int_0^n x^2 \mathrm{d} x= \left. \frac{1}{3}x^3 \right|_0^n = \frac{1}{3}n^3$

所以：

$\begin{aligned} \sum_{i=1}^n i^2 &= \frac{1}{3}n^3 + \frac{n(n+1)}{2} - \frac{n}{3} \\ &= \frac{n(n+1)(2n+1)}{6} \\ \end{aligned}$

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