Determinants and equalities
Posted: Thu Jun 09, 2016 9:18 pm
Consider the real numbers \( a_{ij} , \; i=1,2, \dots,n-2 \; , \;\; j=1, 2, \dots, n \;\;\; n \geq 3 \) and the determinants
A_k = $$ \begin{vmatrix}
1 &\cdots &1 &1 &\cdots &1 \\
a_{11}&\cdots &a_{1, k-1} &a_{1, k+1} &\cdots & a_{1n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{n-2, 1} &\cdots &a_{n-2, k-1} &a_{n-2, k+1} &\cdots & a_{n-2, n}
\end{vmatrix}$$
Prove that:
a. \( A_1+A_3+A_5+\cdots=A_2+A_4+A_6+\cdots \).
b. For any positive integer \( n \geq 3 \) the following identity holds: \( \displaystyle \sum_{k=1}^{n}\frac{\left ( -1 \right )^k k^2}{\left ( n-k \right )!\left ( n+k \right )!}=0 \) .
D. Andrica
A_k = $$ \begin{vmatrix}
1 &\cdots &1 &1 &\cdots &1 \\
a_{11}&\cdots &a_{1, k-1} &a_{1, k+1} &\cdots & a_{1n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{n-2, 1} &\cdots &a_{n-2, k-1} &a_{n-2, k+1} &\cdots & a_{n-2, n}
\end{vmatrix}$$
Prove that:
a. \( A_1+A_3+A_5+\cdots=A_2+A_4+A_6+\cdots \).
b. For any positive integer \( n \geq 3 \) the following identity holds: \( \displaystyle \sum_{k=1}^{n}\frac{\left ( -1 \right )^k k^2}{\left ( n-k \right )!\left ( n+k \right )!}=0 \) .
D. Andrica