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

发布时间：2023年12月27日

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）。

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

首先看常规解法（即不用组合数学的方法）:
在这里插入图片描述

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

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

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

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

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

文章来源:https://blog.csdn.net/RK_Dangerous/article/details/135120261
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