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

2023-12-27 06:17:19

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?)
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):
在这里插入图片描述


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.
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 ) = 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) 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)
2 ( C ( 1 , 2 ) + C ( 2 , 2 ) + ? ? ? + C ( n , 2 ) ) = 2 ( n C ( 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 ( 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)-(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?

文章来源:https://blog.csdn.net/RK_Dangerous/article/details/135120261
本文来自互联网用户投稿,该文观点仅代表作者本人,不代表本站立场。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。