Q85 / 200 · Manipal (MET) 2014 · Mathematics
If $\quad \Delta_r=\left|\begin{array}{ccc}1 & n & n \\ 2 r & n^2+n+1 & n^2+n \\ 2 r-1 & n^2 & n^2+n+1\end{array}\right| \quad$ and $\sum_{r=1}^n \Delta_r=56$, then $n$ is equal to
- A4
- B6
- Correct answerC7
- D8
Solution
Putting $r=1,2,3, \ldots, n$ and using the formula
$\begin{aligned}
& \Sigma 1=n \\
& \text { and } \quad \Sigma r=\frac{n(n+1)}{2} \\
& \Sigma(2 r-1)=1+3+5+\ldots=n^2 \\
& \therefore \sum_{r=1}^n \Delta_r=\left|\begin{array}{cc}
n & n \\
n(n+1) & n^2+n+1 \\
n^2 & n^2
\end{array} c \begin{array}{l}
n^2+n \\
n^2+n+1
\end{array}\right|=56
\end{aligned}$
Applying $C_1 \rightarrow C_1-C_3, C_2 \rightarrow C_2-C_3$, we get $\begin{aligned} &\left|\begin{array}{ccc} 0 & 0 & n \\ 0 & 1 & n^2+n \\ -n-1 & -n-1 & n^2+n+1 \end{array}\right|=56 \\ & \Rightarrow \\ & \Rightarrow n(n+1)=56 \\ & \Rightarrow \\ & n^2+n-56=0 \end{aligned}$ $\begin{aligned} \Rightarrow & & (n+8)(n-7) & =0 \\ \Rightarrow & & n & =7,(n \neq-8)\end{aligned}$
Applying $C_1 \rightarrow C_1-C_3, C_2 \rightarrow C_2-C_3$, we get $\begin{aligned} &\left|\begin{array}{ccc} 0 & 0 & n \\ 0 & 1 & n^2+n \\ -n-1 & -n-1 & n^2+n+1 \end{array}\right|=56 \\ & \Rightarrow \\ & \Rightarrow n(n+1)=56 \\ & \Rightarrow \\ & n^2+n-56=0 \end{aligned}$ $\begin{aligned} \Rightarrow & & (n+8)(n-7) & =0 \\ \Rightarrow & & n & =7,(n \neq-8)\end{aligned}$
Hence, the correct answer is option: 'C'.