Question
If \[{\Delta _r} = \left| {\begin{array}{*{20}{c}}
{r - 1}&n&6\\
{{{\left( {r - 1} \right)}^2}}&{2{n^2}}&{4n - 2}\\
{{{\left( {r - 1} \right)}^2}}&{3{n^3}}&{3{n^2} - 3n}
\end{array}} \right|,\] then $$\sum\limits_{r = 1}^n {{\Delta _r}} $$ is
A.
$$0$$
B.
$$1$$
C.
$$3$$
D.
$$ - 1$$
Answer :
$$0$$
Solution :
Since $$C_1$$ has variable terms and $$C_2$$ and $$C_3$$ are constant, summation runs on $$C_1 .$$ Therefore,
\[\begin{array}{l}
\sum\limits_{r = 1}^n {{\Delta _r}} = \left| {\begin{array}{*{20}{c}}
{\sum\limits_1^n {\left( {r - 1} \right)} }&n&6\\
{\sum\limits_1^n {{{\left( {r - 1} \right)}^2}} }&{2{n^2}}&{4n - 2}\\
{\sum\limits_1^n {{{\left( {r - 1} \right)}^3}} }&{3{n^3}}&{3{n^2} - 3n}
\end{array}} \right|\\
= \,\left| {\begin{array}{*{20}{c}}
{\frac{1}{2}\left( {n - 1} \right)n}&n&6\\
{\frac{1}{6}\left( {n - 1} \right)n\left( {2n - 1} \right)}&{2{n^2}}&{4n - 2}\\
{\frac{1}{4}{{\left( {n - 1} \right)}^2}{n^2}}&{3{n^3}}&{3{n^2} - 3n}
\end{array}} \right|
\end{array}\]
Taking $$\frac{1}{{12}}n\left( {n - 1} \right)$$ common from $$C_1$$ and $$n$$ common from $$C_2 ,$$ we get
\[\sum {{\Delta _r} = \frac{1}{{12}}{n^2}\left( {n - 1} \right) \times \left| {\begin{array}{*{20}{c}}
6&1&6\\
{2\left( {2n - 1} \right)}&{2n}&{2\left( {2n - 1} \right)}\\
{3n\left( {n - 1} \right)}&{3{n^3}}&{3n\left( {n - 1} \right)}
\end{array}} \right|} \]
$$ = \,0\,\,\,\left[ {\because \,{C_1}\,{\text{and}}\,{C_3}\,{\text{are}}\,{\text{identical}}} \right]$$