Question

The value of $\sum_{n = 1}^{N}U_{n}$, if $U_{n} = \left| \begin{matrix} n & 1 & 5 \\ n^{2} & 2N + 1 & 2N + 1 \\ n^{3} & 3N^{2} & 3N \end{matrix} \right|$ is

• A. 0
• B. 1
• C. –1
• D. None of these

Answer 1

Answer: (A)

0

Answer 2

\frac{N(N + 1)}{2} & 1 & 5 \\ \frac{N(N + 1)(2N + 1)}{6} & 2N + 1 & 2N + 1 \\ \left\{ \frac{N(N + 1)}{2} \right\}^{2} & 3N^{2} & 3N \end{matrix} \right|} = \frac{N(N + 1)}{12}\left| \begin{matrix} 6 & 1 & 5 \\ 4N + 2 & 2N + 1 & 2N + 1 \\ 3N(N + 1) & 3N^{2} & 3N \end{matrix} \right|$$ Applying $C_{3} \rightarrow C_{3} + C_{2}$ $= \frac{N(N + 1)}{12}\left| \begin{matrix} 6 & 1 & 6 \\ 4N + 2 & 2N + 1 & 4N + 2 \\ 3N(N + 1) & 3N^{2} & 3N(N + 1) \end{matrix} \right|$ = 0 [$\because C_{1}$ and $C_{3}$ are identical]