计算 det[1^2,2^2,3^2,…,n^2；n^2,1^2,2^2,…,(n-1)^2；…；2^2,3^2,4^2,…,1^2]

wilsony

谁能算出？

wilsony

陆老师能算出这个行列式吗？

luyuanhong

wilsony

多谢陆老师！

fungarwai

\(q=e^{\frac{2\pi j}{n}i},~p(k)=k^2,~\Delta p(k)=2k+1,~\Delta^2 p(k)=2\)

\(\displaystyle f(n)=\frac{n^2}{q-1}-\frac{2n+1}{(q-1)^2}+\frac{2}{(q-1)^3}q\)

\(\displaystyle \sum_{k=0}^{n-1}(k+1)^2 q^k
=\sum_{k=1}^n k^2 q^{k-1}=f(n)q^n-f(0)
=\frac{n^2}{q-1}-\frac{2n}{(q-1)^2}~(q\neq 1)\)

\(\displaystyle \sum_{k=0}^{n-1}(k+1)^2=\binom{n}{1}+3\binom{n}{2}+2\binom{n}{3}\)

\(\displaystyle \prod_{j=0}^{n-1}\sum_{k=0}^{n-1}(k+1)^2 q^k
=\left(\binom{n+1}{2}+2\binom{n+1}{3}\right)
\prod_{j=1}^{n-1} \frac{n^2}{(q-1)^2}\left(q-1-\frac{2}{n}\right)\)
\(\displaystyle =\frac{n(n+1)(2n+1)}{6}\cdot
\frac{n^{2(n-1)}(-1)^n\left(\left(1+\frac{2}{n}\right)^n-1\right)}
{n^2(-\frac{2}{n})}\)
\(\displaystyle =\frac{(n+1)(2n+1)}{12}\cdot
n^{2n-2}(-1)^{n-1}\left(\left(1+\frac{2}{n}\right)^n-1\right)\)
\(\displaystyle =\frac{(-1)^{n-1}}{12}n^{n-2}(n+1)(2n+1)\left((n+2)^n-n^n\right)\)

luyuanhong

楼上 fungarwai 的解答很好！已收藏。

luyuanhong

楼上 fungarwai 的解答很好！已收藏。