Question

If U_n = $\left| \begin{matrix} 1 & k & k \\ 2n & k^{2} + k + 1 & k^{2} + k \\ 2n - 1 & k^{2} & k^{2} + k + 1 \end{matrix} \right|$ and

$\sum_{n = 1}^{k}{U_{n} = 72}$then k =

• A. 8
• B. 9
• C. 6
• D. None of these

Answer 1

Answer: (A)

8

Answer 2

We have,

$\sum_{n = 1}^{k}U_{n}$= $\left| \begin{matrix} \sum_{n = 1}^{k}1 & k & k \\ 2\sum_{n = 1}^{k}n & k^{2} + k + 1 & k^{2} + k \\ 2\sum_{n = 1}^{k}n - \sum_{n = 1}^{k}1 & k^{2} & k^{2} + k + 1 \end{matrix} \right|$

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

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

[Applying C₂ → C₂ – C₁]

= k(k² + k + 1) – k³ = k(k + 1) = 72 (given)

⇒ k = 8

Hence (1) is correct answer.