Question

If ${△}_r=\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r-1& n^2& n^2+n+1\end{array}\right|$ and $\sum _{r=0}^m{△}_r=56$, then $n=$

• A. $4$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C

$7$

Step 1. Find the value of $n$:

Given,

${△}_r=\left|\begin{array}{ccc}1& n& n\\ 2r& n^2+n+1& n^2+n\\ 2r-1& n^2& n^2+n+1\end{array}\right|$ and ${\displaystyle \sum _{r=0}^n}{△}_r=56$

If we put $r=1,2,3,4,....n$, we get

$\sum _{r=1}^n{△}_r=\left|\begin{array}{ccc}n& n& n\\ n(n+1)& n^2+n+1& n^2+n\\ n^2& n^2& n^2+n+1\end{array}\right|$ $\left[\mathbf{\because }\mathbf{\sum }_{}\mathbf{1}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}\mathbf{;}\mathbf{}\mathbf{\sum }_{}\mathit{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}}{\mathbf{2}}\mathbf{;}\mathbf{}\mathbf{\sum }_{}\left(2r-1\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{+}\mathbf{}\mathbf{5}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{}{\mathit{n}}^{\mathbf{2}}\right]$

$\begin{array}{rcl}\sum _{r=1}^n{△}_r& =& \left|\begin{array}{ccc}n& n& n\\ n(n+1)& n^2+n+1& n^2+n\\ n^2& n^2& n^2+n+1\end{array}\right|\\ & =& 56\end{array}$

Step 2: By doing column transformation as $C_1\to C1-C_3$ and $C_2\to C_2-C_3$, we get

$\begin{array}{rcl}\left|\begin{array}{ccc}0& 0& n\\ 0& 1& n^2+n\\ n-1& -n-1& n^2+n+1\end{array}\right|& =& 56\\ & & \end{array}$

$\Rightarrow$ $n(n+1)=56$

$\Rightarrow$ $n^2+n=56$

$\Rightarrow$ $n^2+n-56=0$

$\Rightarrow$ $n^2+8n-7n-56=0$

$\Rightarrow$ $\left(n+8\right)\left(n-7\right)=0$

$\Rightarrow$ $n=7,-8$ (but $n$ cannot be negative)

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{7}$

Hence, Option ‘C’ is Correct.