A) 0
B) 1
C) - 1
D) None of these
Correct Answer: A
Solution :
\[\sum\limits_{n=1}^{N}{{{U}_{n}}=}\left| \,\begin{matrix} \frac{N(N+1)}{2} & 1 & 5 \\ \frac{N(N+1)(2N+1)}{6} & 2N+1 & 2N+1 \\ {{\left\{ \frac{N(N+1)}{2} \right\}}^{2}} & 3{{N}^{2}} & 3N \\ \end{matrix}\, \right|\] \[=\frac{N(N+1)}{12}\left| \,\begin{matrix} 6 & 1 & 5 \\ 4N+2 & 2N+1 & 2N+1 \\ 3N(N+1) & 3{{N}^{2}} & 3N \\ \end{matrix}\, \right|\] \[=\left| \,\begin{matrix} 6 & 1 & 6 \\ 4N+2 & 2N+1 & 4N+2 \\ 3N(N+1) & 3{{N}^{2}} & 3N(N+1) \\ \end{matrix}\, \right|=0\], {Applying\[{{C}_{3}}\to {{C}_{3}}+{{C}_{2}}\}\].You need to login to perform this action.
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