【组合数学题解】利用m2=2C(m,2)+C(m,1)求12+22+···+n2的值
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
(
m
2
)
+
(
m
1
)
m^2=2 \left( \begin{array}{lc} m\\ 2 \end{array} \right) + \left( \begin{array}{lc} m\\ 1 \end{array} \right)
m2=2(m2?)+(m1?)
求
1
2
+
2
2
+
?
?
?
+
n
2
1^2+2^2+···+n^2
12+22+???+n2的值.
首先看常规解法（即不用组合数学的方法）:
解：
step1：先将
m
2
m^2
m2对应的组合式代入
1
2
+
2
2
+
?
?
?
+
n
2
1^2+2^2+···+n^2
12+22+???+n2.
1
2
+
2
2
+
?
?
?
+
n
2
=
2
(
1
2
)
+
(
1
1
)
+
2
(
2
2
)
+
(
2
1
)
+
2
(
3
2
)
+
(
3
1
)
+
?
?
?
+
2
(
n
2
)
+
(
n
1
)
=
2
(
(
1
2
)
+
(
2
2
)
+
?
?
?
+
(
n
2
)
)
+
(
(
1
1
)
+
(
2
1
)
+
?
?
?
+
(
n
1
)
)
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)
12+22+???+n2=2(12?)+(11?)+2(22?)+(21?)+2(32?)+(31?)+???+2(n2?)+(n1?)=2((12?)+(22?)+???+(n2?)?)+((11?)+(21?)+???+(n1?)?)
step2：对于后半部分
(
(
1
1
)
+
(
2
1
)
+
?
?
?
+
(
n
1
)
)
\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)
((11?)+(21?)+???+(n1?)?)整理.
(
(
1
1
)
+
(
2
1
)
+
?
?
?
+
(
n
1
)
)
=
1
+
2
+
?
?
?
+
n
=
(
n
+
1
)
n
2
\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}
((11?)+(21?)+???+(n1?)?)=1+2+???+n=2(n+1)n?
step3：对于前半部分
2
(
(
1
2
)
+
(
2
2
)
+
?
?
?
+
(
n
2
)
)
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)
2((12?)+(22?)+???+(n2?)?)整理.
由
C
(
n
,
k
)
=
C
(
n
?
1
,
k
?
1
)
+
C
(
n
?
1
,
k
)
C(n,k)=C(n-1,k-1)+C(n-1,k)
C(n,k)=C(n?1,k?1)+C(n?1,k)得：
C
(
1
,
2
)
=
C
(
1
,
2
)
C(1, 2)=C(1,2)
C(1,2)=C(1,2)
C
(
2
,
2
)
=
C
(
1
,
1
)
+
C
(
1
,
2
)
C(2,2)=C(1,1)+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(3,2)=C(2,1)+C(2,2)=C(2,1)+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
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=
C
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n
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1
,
1
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+
C
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n
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1
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=
C
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1
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C
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2
,
1
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?
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+
C
(
2
,
1
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+
C
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1
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1
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C
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1
,
2
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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)
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
(
n
C
(
1
,
2
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+
(
n
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1
)
C
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1
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1
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+
(
n
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2
)
C
(
2
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1
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+
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+
(
n
?
(
n
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1
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)
(
C
(
n
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1
,
1
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+
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n
?
n
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C
(
n
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1
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)
=
2
(
n
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0
+
(
n
?
1
)
?
1
+
(
n
?
2
)
?
2
+
?
?
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+
(
n
?
(
n
?
1
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)
?
(
n
?
1
)
+
(
n
?
n
)
?
n
)
=
2
(
(
0
?
n
+
1
?
n
+
2
?
n
+
?
?
?
+
n
?
n
)
?
(
1
2
+
2
2
+
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?
?
+
n
2
)
)
=
2
(
n
(
n
(
n
+
1
)
2
)
?
(
1
2
+
2
2
+
?
?
?
+
n
2
)
)
=
n
2
(
n
+
1
)
?
2
(
1
2
+
2
2
+
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+
n
2
)
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)
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)?(12+22+???+n2))=2(n(2n(n+1)?)?(12+22+???+n2))=n2(n+1)?2(12+22+???+n2)
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
)
+
(
n
+
1
)
n
2
1^2+2^2+···+n^2=n^2(n+1)-2(1^2+2^2+···+n^2)+\frac{(n+1)n}{2}
12+22+???+n2=n2(n+1)?2(12+22+???+n2)+2(n+1)n?
step5：将等号右边的
1
2
+
2
2
+
?
?
?
+
n
2
1^2+2^2+···+n^2
12+22+???+n2移到左边，并使系数为1，整理:
3
(
1
2
+
2
2
+
?
?
?
+
n
2
)
=
n
2
(
n
+
1
)
+
(
n
+
1
)
n
2
3(1^2+2^2+···+n^2)=n^2(n+1)+\frac{(n+1)n}{2}
3(12+22+???+n2)=n2(n+1)+2(n+1)n?
1
2
+
2
2
+
?
?
?
+
n
2
=
1
3
(
n
2
(
n
+
1
)
+
(
n
+
1
)
n
2
)
=
(
2
n
+
1
)
(
n
+
1
)
n
6
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}
12+22+???+n2=31?(n2(n+1)+2(n+1)n?)=6(2n+1)(n+1)n?
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