Chủ đề này có 2 trả lời

[Nobodyv3]

Tính giá trị của
$$\displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$$

[perfectstrong]

Tính giá trị của
$$\displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$$

\[\begin{gathered}
  F\left( {n,3} \right) = \mathop {\sum \sum \sum }\limits_{1 \leqslant i \leqslant j \leqslant k \leqslant n} ijk \hfill \\
  F\left( {n,2} \right) = \mathop {\sum \sum }\limits_{1 \leqslant i \leqslant j \leqslant n} ij \hfill \\
  F\left( {n,1} \right) = \mathop \sum \limits_{1 \leqslant i \leqslant n} i = \frac{{n\left( {n + 1} \right)}}{2} \hfill \\
  F\left( {n + 1,3} \right) = {\left( {n + 1} \right)^3} + {\left( {n + 1} \right)^2}F\left( {n,1} \right) + \left( {n + 1} \right)F\left( {n,2} \right) + F\left( {n,3} \right) \hfill \\
  F\left( {n + 1,2} \right) = {\left( {n + 1} \right)^2} + \left( {n + 1} \right)F\left( {n,1} \right) + F\left( {n,2} \right) \hfill \\
\end{gathered} \]

[Nobodyv3]

@perfecstrong:Cám ơn anh đã khui lại!

Ta có :
$\begin {align*}
\underset{1\le i\le j\le k\le n}{\mathop{\sum }}\,ijk&=\underset{1\le i< j< k\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le i= j< k\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le i< j= k\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le i= j= k\le n}{\mathop{\sum }}\,ijk\\
&=\underset{1\le i< j< k\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le i= j< k\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le k< i= j\le n}{\mathop{\sum }}\,ijk\quad +\underset{1\le i= j= k\le n}{\mathop{\sum }}\,ijk\\
&=\underset{1\le i< j< k\le n}{\mathop{\sum }}\,ijk\quad+\underset{1\le i= j, k\le n}{\mathop{\sum }}\,ijk
\quad (*) \\
&=\frac {1}{48}n^2(n+1)^2(n-1)(n-2)+\frac {1}{12}n^2(n+1)^2(2n+1)\\
&=\frac {1}{48}n^2(n+1)^2(n^2+5n+6)\\&=\boldsymbol {\frac {1}{48}n^2(n+1)^2(n+2)(n+3)}
\end{align*}$

$\boldsymbol {(*)}$ Ghi chú :
$\bullet $ Chứng minh $\underset{1\le i< j< k\le n}{\mathop{\sum }}\,ijk=\frac {1}{48}n^2(n+1)^2(n-1)(n-2)$ xin xem tại đây https://diendantoanh...eq-ijkleq-nijk/
$\bullet \underset{1\le i= j, k\le n}{\mathop{\sum }}\,ijk=\left ( \sum_{j=1}^{n}j^2 \right )\left ( \sum_{k=1}^{n}k \right )=\frac {1}{6}n(n+1)(2n+1)\cdot \frac {1}{2}n(n+1)=\frac {1}{12}n^2(n+1)^2(2n+1)$

Cách khác : Áp dụng bổ đề Burnside, ta có :
TH 1: Các biến chạy $i,j,k$ không có ràng buộc gì.
${{S}_{1}}=\sum\limits_{1\le i,j,k\le n}{ijk}=\left( \sum\limits_{i=1}^{n}{i} \right)\left( \sum\limits_{j=1}^{n}{j} \right)\left( \sum\limits_{k=1}^{n}{k} \right)={{\left[ \frac{1}{2}n\left( n+1 \right) \right]}^{3}}=\frac{1}{8}{{n}^{3}}{{\left( n+1 \right)}^{3}}$
TH 2: Có 2 trong 3 biến $i,j,k$ bằng nhau.
${{S}_{2}}=\sum\limits_{1\le i=j,k\le n}{ijk}=\sum\limits_{1\le j,k\le n}{{{j}^{2}}k=}\left( \sum\limits_{j=1}^{n}{{{j}^{2}}} \right)\left( \sum\limits_{k=1}^{n}{k} \right)$

${{S}_{2}}=\left[ \frac{1}{6}n\left( n+1 \right)\left( 2n+1 \right) \right]\left[ \frac{1}{2}n\left( n+1 \right) \right]=\frac{1}{12}{{n}^{2}}{{\left( n+1 \right)}^{2}}\left( 2n+1 \right)$

TH 3: Cả 3 biến bằng nhau $ i=j=k$
${{S}_{3}}=\sum\limits_{1\le i=j=k\le n}{ijk}=\sum\limits_{i=1}^{n}{{{i}^{3}}}=\frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}}$

Theo bổ đề Burnside ta có :
$\sum\limits_{1\le i\le j\le k\le n}{ijk}=\frac{1}{3!}\left( {{S}_{1}}+3{{S}_{2}}+2{{S}_{3}} \right)=\frac{1}{6}\left[ \frac{1}{8}{{n}^{3}}{{\left( n+1 \right)}^{3}}+3\cdot \frac{1}{12}{{n}^{2}}{{\left( n+1 \right)}^{2}}\left( 2n+1 \right)+2\cdot \frac{1}{4}{{n}^{2}}{{\left( n+1 \right)}^{2}} \right]$
$\Rightarrow \boldsymbol {\sum\limits_{1\le i\le j\le k\le n}{ijk}=\frac{1}{48}{{n}^{2}}{{\left( n+1 \right)}^{2}}\left( n+2 \right)\left( n+3 \right)}$

Bài viết đã được chỉnh sửa nội dung bởi Nobodyv3: 16-02-2023 - 00:35