A) \[0\]
B) \[1\]
C) \[3\]\[-1\]
D)
Correct Answer: A
Solution :
[a] Since \[{{C}_{1}}\] has variable terms and \[{{C}_{2}}\] and \[{{C}_{3}}\] are constant, summation runs on \[{{C}_{1}}\]. Therefore, \[\sum\limits_{r=1}^{n}{{{\Delta }_{r}}}=\left| \begin{matrix} \sum\limits_{1}^{n}{(r-1)} & n & 6 \\ \sum\limits_{1}^{n}{{{(r-1)}^{2}}} & 2{{n}^{2}} & 4n-2 \\ \sum\limits_{1}^{n}{{{(r-1)}^{3}}} & 3{{n}^{2}} & 3{{n}^{2}}-3n \\ \end{matrix} \right|\] \[=\left| \begin{matrix} \frac{1}{2}(n-1)n & n & 6 \\ \frac{1}{6}(n-1)n(2n-1) & 2{{n}^{2}} & 4n-2 \\ \frac{1}{4}{{(n-1)}^{2}}{{n}^{2}} & 3{{n}^{3}} & 3{{n}^{2}}-3n \\ \end{matrix} \right|\] Taking \[\frac{1}{12}n(n-1)\] common from \[{{C}_{1}}\] and n common from \[{{C}_{2}}\], we get \[\sum{{{\Delta }_{r}}=\frac{1}{12}{{n}^{2}}(n-1)\times \left| \begin{matrix} 6 & 1 & 6 \\ 2(2n-1) & 2n & 2(2n-1) \\ 3n(n-1) & 3{{n}^{2}} & 3n(n-1) \\ \end{matrix} \right|}\] \[=0[\because \,\,\,\,{{C}_{1}}\,\,and\,\,{{C}_{3}}\,\,are\,\,identical]\]
You need to login to perform this action.
You will be redirected in
3 sec
