Question

If D_r = $\left| \begin{matrix} r - 1 & n & 6 \\ (r - 1)^{2} & 2n^{2} & 4n - 2 \\ (r - 1)^{3} & 3n^{2} & 3n^{2} - 3n \end{matrix} \right|$ then $\sum_{r = 1}^{n}\Delta_{r}$equals-

• A. n²(n + 1)
• B. n(n + 1)²
• C. $\frac{1}{12}$n(n³ + 1)
• D. None

Answer 1

Answer: (D)

None

Answer 2

Using the sum property of determinants, we get

\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{matrix} \right|}$$ But $\sum_{r = 1}^{n}{(r - 1)}$=$\frac{1}{2}$ n (n – 1), $\sum_{r = 1}^{n}{(r - 1)^{2}}$= $\frac{1}{6}$n (n – 1) (2n – 1) and $\sum_{r = 1}^{n}{(r - 1)^{3}}$=$\frac{1}{4}$n² (n – 1)².Thus, $\sum_{r = 1}^{n}\Delta_{r}$=$\left| \begin{matrix} \frac{1}{2}n(n - 1) & n & 6 \\ \frac{1}{6}n(n - 1)(2n - 1) & 2n^{2} & 4n - 2 \\ \frac{1}{4}n^{2}(n - 1)^{2} & 3n^{3} & 3n^{2} - 3n \end{matrix} \right|$ Taking $\frac{1}{12}$n (n – 1) common from C₁ and n from C₂, we get $\sum_{r = 1}^{n}\Delta_{r}$=$\frac{1}{12}$n (n – 1) (n) $$\left| \begin{matrix} 6 & 1 & 6 \\ 2(2n - 1) & 2n & 4n - 2 \\ 3n(n - 1) & 3n^{2} & 3n^{2} - 3n \end{matrix} \right|$$ = $\frac{1}{12}$ n (n – 1) (n) (0) = 0 [Q C₁ and C₃ are identical]