【组合数学题解】利用m²=2C(m,2)+C(m,1)求1²+2²+···+n²的值

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这道题我在网上没有找到满意的解答，自己也懒得找教材有没有配套的答案，所以把解题过程记录在这里了（没有和老师对答案，不管是不是出题人想要的过程反正是利用组合数学的知识求出来了）。

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题目：
利用$$m^2=2\left(\begin{array}{l}m\\ 2\end{array}\right)+\left(\begin{array}{l}m\\ 1\end{array}\right)$$
求$1^2+2^2+\cdot \cdot \cdot +n^2$的值.

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首先看常规解法（即不用组合数学的方法）:

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解：
step1：先将$m^2$对应的组合式代入$1^2+2^2+\cdot \cdot \cdot +n^2$.
$$\begin{gathered} 1^2+2^2+\cdot \cdot \cdot +n^2=2\left(\begin{array}{l}1\\ 2\end{array}\right)+\left(\begin{array}{l}1\\ 1\end{array}\right)+2\left(\begin{array}{l}2\\ 2\end{array}\right)+\left(\begin{array}{l}2\\ 1\end{array}\right)+2\left(\begin{array}{l}3\\ 2\end{array}\right)+\left(\begin{array}{l}3\\ 1\end{array}\right)+\cdot \cdot \cdot +2\left(\begin{array}{l}n\\ 2\end{array}\right)+\left(\begin{array}{l}n\\ 1\end{array}\right) \\ =2\left(\begin{array}{l}\left(\begin{array}{l}1\\ 2\end{array}\right)+\left(\begin{array}{l}2\\ 2\end{array}\right)+\cdot \cdot \cdot +\left(\begin{array}{l}n\\ 2\end{array}\right)\end{array}\right)+\left(\begin{array}{l}\left(\begin{array}{l}1\\ 1\end{array}\right)+\left(\begin{array}{l}2\\ 1\end{array}\right)+\cdot \cdot \cdot +\left(\begin{array}{l}n\\ 1\end{array}\right)\end{array}\right) \end{gathered}$$

step2：对于后半部分$\left(\begin{array}{l}\left(\begin{array}{l}1\\ 1\end{array}\right)+\left(\begin{array}{l}2\\ 1\end{array}\right)+\cdot \cdot \cdot +\left(\begin{array}{l}n\\ 1\end{array}\right)\end{array}\right)$整理.
$$\begin{gathered} \left(\begin{array}{l}\left(\begin{array}{l}1\\ 1\end{array}\right)+\left(\begin{array}{l}2\\ 1\end{array}\right)+\cdot \cdot \cdot +\left(\begin{array}{l}n\\ 1\end{array}\right)\end{array}\right)=1+2+\cdot \cdot \cdot +n \\ =\frac{(n+1)n}{2} \end{gathered}$$

step3：对于前半部分$2\left(\begin{array}{l}\left(\begin{array}{l}1\\ 2\end{array}\right)+\left(\begin{array}{l}2\\ 2\end{array}\right)+\cdot \cdot \cdot +\left(\begin{array}{l}n\\ 2\end{array}\right)\end{array}\right)$整理.
由$C(n,k)=C(n-1,k-1)+C(n-1,k)$得：
$C(1,2)=C(1,2)$
$C(2,2)=C(1,1)+C(1,2)$
$C(3,2)=C(2,1)+C(2,2)=C(2,1)+C(1,1)+C(1,2)$
$\cdot \cdot \cdot$
$$C(n,2)=C(n-1,1)+C(n-1,2)=C(n-1,1)+C(n-2,1)\cdot \cdot \cdot +C(2,1)+C(1,1)+C(1,2)$$
则$$\begin{gathered} 2(C(1,2)+C(2,2)+\cdot \cdot \cdot +C(n,2)) \\ =2(nC(1,2)+(n-1)C(1,1)+(n-2)C(2,1)+\cdot \cdot \cdot +(n-(n-1))(C(n-1,1)+(n-n)C(n,1)) \\ =2(n\ast 0+(n-1)\ast 1+(n-2)\ast 2+\cdot \cdot \cdot +(n-(n-1))\ast (n-1)+(n-n)\ast n) \\ =2((0\ast n+1\ast n+2\ast n+\cdot \cdot \cdot +n\ast n)-(1^2+2^2+\cdot \cdot \cdot +n^2)) \\ =2(n(\frac{n(n+1)}{2})-(1^2+2^2+\cdot \cdot \cdot +n^2)) \\ =n^2(n+1)-2(1^2+2^2+\cdot \cdot \cdot +n^2) \end{gathered}$$

step4：将第2步和第3步得出的式子代入原式.
$1^2+2^2+\cdot \cdot \cdot +n^2=n^2(n+1)-2(1^2+2^2+\cdot \cdot \cdot +n^2)+\frac{(n+1)n}{2}$

step5：将等号右边的$1^2+2^2+\cdot \cdot \cdot +n^2$移到左边，并使系数为1，整理:
$3(1^2+2^2+\cdot \cdot \cdot +n^2)=n^2(n+1)+\frac{(n+1)n}{2}$
$$\begin{gathered} 1^2+2^2+\cdot \cdot \cdot +n^2=\frac{1}{3}(n^2(n+1)+\frac{(n+1)n}{2}) \\ =\frac{(2n+1)(n+1)n}{6} \end{gathered}$$

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