Chủ đề này có 3 trả lời

[stuart clark]

Finding $\displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$

[perfectstrong]

Finding $\displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$

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

Ta có hệ thức truy hồi:

\[\begin{gathered}
  F\left( {n + 1,3} \right) = F\left( {n,3} \right) + \left( {n + 1} \right)F\left( {n,2} \right) \hfill \\
  F\left( {n + 1,2} \right) = F\left( {n,2} \right) + \left( {n + 1} \right)F\left( {n,1} \right) \hfill \\
\end{gathered} \]

Từ đó ...

[Nobodyv3]

Kết quả của mình là
$$ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk=\frac {(n-2)(n-1)n^2(n+1)^2}{48}$$

[Nobodyv3]

Để đỡ phải đánh máy, ta ký hiệu : Tính
$$ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk=\sum_{i,j,k=1\atop (i<j<k)}^{n}ijk$$
Giải :
Ta thấy :
$\begin {align*}
&\sum_{i,j,k=1\atop (i<j<k)}^{3}ijk=6;\:\sum_{i,j,k=1\atop (i<j<k)}^{4}ijk=50=6+44=\sum_{i,j,k=1\atop (i<j<k)}^{3}ijk+4\sum_{i,j=1\atop (i<j)}^{3}ij\\
&\sum_{i,j,k=1\atop (i<j<k)}^{5}ijk=225=50+175=\sum_{i,j,k=1\atop (i<j<k)}^{4}ijk+5\sum_{i,j=1\atop (i<j)}^{4}ij\\
&......\\
&\text{Bằng quy nạp, ta chứng minh được với $n>3$:}\\
&\sum_{i,j,k=1\atop (i<j<k)}^{n}ijk= \sum_{i,j,k=1\atop (i<j<k)}^{n-1}ijk+n\sum_{i,j=1\atop (i<j)}^{n-1}ij\\
&\text{Suy ra:}\\
&\sum_{i,j,k=1\atop (i<j<k)}^{n}ijk- \sum_{i,j,k=1\atop (i<j<k)}^{n-1}ijk=n\sum_{i,j=1\atop (i<j)}^{n-1}ij=n\cdot \frac {(n-2)(n-1)n(3n-1)}{24}=\frac {3n^5-10n^4+9n^3-2n^2}{24}
\end {align*}$
Thay $ n=4,5,6,...,n$ và cộng lại :
$\begin {align*}
\sum_{i,j,k=1\atop (i<j<k)}^{n}ijk- \sum_{i,j,k=1\atop (i<j<k)}^{3}ijk&=\frac {1}{24} \sum_{i=4}^{n}\left (3i^5-10i^4+9i^3-2i^2  \right ) \\
&=\frac {1}{24} \sum_{i=1}^{n}\left (3i^5-10i^4+9i^3-2i^2  \right )\\
&-\frac {1}{24}\left [ 3\left ( 1^5+2^5+3^5 \right )-10\left ( 1^4+2^4+3^4 \right )
+9 \left ( 1^3+2^3+3^3 \right )-2\left ( 1^2+2^2+3^2 \right ) \right ]\\
\text {Do đó với $n\geq 3$ ta có :}\\
\sum_{i,j,k=1\atop (i<j<k)}^{n}ijk&=\frac {1}{24} \sum_{i=1}^{n}\left (3i^5-10i^4+9i^3-2i^2  \right )\\
&=\frac {1}{24}\left [3\left ( \frac {n^6}{6}+\frac {n^5}{2}+\frac {5n^4}{12}-\frac{n^2}{12} \right )-10\left ( \frac {n^5}{5}+\frac {n^4}{2}+\frac {n^3}{3}-\frac{n}{30} \right )+9\left ( \frac {n^4}{4}+\frac {n^3}{2}+\frac{n^2}{4} \right )-2 \left ( \frac {n^3}{3}+\frac {n^2}{2}+\frac {n}{6}\right ) \right ]\\
&=\frac {1}{48}\left (n^6-n^5-3n^4+n^3+2n^2  \right )\\
&=\boldsymbol {\frac {(n-2)(n-1)n^2(n+1)^2}{48}}
\end {align*}$