Question

Let ${\Delta }_r=\left|\begin{array}{ccc}r-1& n& 6\\ (r-1{)}^2& 2n^2& 4n-2\\ (r-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$. Then the value of $\sum _{r=1}^n{\Delta }_r$ is:

• A. $1$
• B. $3$
• C. $2$
• D. $0$

Answer 1

The correct option is D $0$

Since $C_1$ has variable terms and $C_2$ and $C_3$ are constant, summation runs only on $C_1$.
$\therefore \sum _{r=1}^n{\Delta }_r=\left|\begin{array}{ccc}\sum _{r=1}^n(r-1)& n& 6\\ \sum _{r=1}^n(r-1{)}^2& 2n^2& 4n-2\\ \sum _{r=1}^n(r-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$
$=\left|\begin{array}{ccc}\frac{1}{2}(n-1)n& n& 6\\ \frac{1}{6}(n-1)n(2n-1)& 2n^2& 4n-2\\ \frac{1}{4}(n-1{)}^2{n}^2& 3n^3& 3n^2-3n\end{array}\right|$

Taking $\frac{1}{12}n(n-1)$ common from $C_1$ and $n$ from $C_2$, we get
$\sum {\Delta }_r=\frac{1}{12}{n}^2(n-1)\times \left|\begin{array}{ccc}6& 1& 6\\ 2(2n-1)& 2n& 2(2n-1)\\ 3n(n-1)& 3n^2& 3n(n-1)\end{array}\right|$
$\Rightarrow \sum {\Delta }_r=0$, since $C_1$ and $C_3$ are identical.