Question

If $U_n=\left|\begin{array}{ccc}1& k& k\\ 2n& k^2+k+1& k^2+k\\ 2n-1& k^2& k^2+k+1\end{array}\right|$ and ${\sum }_{n=1}^k{U}_n=72$ then $k=$

• A. $8$
• B. $7$
• C. $6$
• D. None of these

Answer 1

The correct option is A $8$

$\sum _{n=1}^k{U}_n=\left|\begin{array}{ccc}{\sum }_{n=1}^{k}1& k& k\\ 2{\sum }_{n=1}^{k}n& k^2+k+1& k^2+k\\ 2{\sum }_{n=1}^{k}n-{\sum }_{n=1}^{k}1& k^2& k^2+k+1\end{array}\right|$

$=\left|\begin{array}{ccc}k& k& k\\ k(k+1)& k^2+k+1& k^2+k\\ k^2& k^2& k^2+k+1\end{array}\right|$

Applying $C_2\to C_2-C_1$

$=\left|\begin{array}{ccc}k& 0& k\\ k^2+k& 1& k^2+k\\ k^2& 0& k^2+k+1\end{array}\right|$

$=k(k^2+k+1)-k^3=k(k+1)=72$ (given)

$\Rightarrow k=8$