Question

If $U_n=\left|\begin{array}{ccc}n& 1& 5\\ n^2& 2N+1& 2N+1\\ n^3& 3N^2& 3N+1\end{array}\right|$ and $\sum _{n=1}^N{U}_n=\lambda \sum _{n=1}^N{n}^2,$ then the value of $\lambda$ is

Answer 1

$\sum _{n=1}^N{U}_n=\left|\begin{array}{ccc}\sum _{n=0}^{N}n& 1& 5\\ \sum _{n=0}^N{n}^2& 2N+1& 2N+1\\ \sum _{n=0}^N{n}^3& 3N^2& 3N+1\end{array}\right|$
$=\frac{N(N+1)}{2}\left|\begin{array}{ccc}1& 1& 5\\ \frac{2N+1}{3}& 2N+1& 2N+1\\ \frac{N(N+1)}{2}& 3N^2& 3N+1\end{array}\right|$
$=\frac{N(N+1)}{2}\cdot \frac{2N+1}{3}\cdot \frac{1}{2}\left|\begin{array}{ccc}1& 1& 5\\ 1& 3& 3\\ N^2+N& 6N^2& 6N+2\end{array}\right|$
$=\frac{1}{2}\sum _{n=1}^N{n}^2\left[\right(N^2+N\left)\right(-12)-6N^2(-2)+(6N+2\left)\right(2\left)\right]$
$=2\sum _{n=1}^N{n}^2\cdots (\because \sum _{n=1}^N{n}^2=\frac{N(N+1)}{2}\cdot \frac{2N+1}{3})$
$\Rightarrow \lambda =2$