Question

Let $△a=\left|\begin{array}{ccc}a-1& n& 6\\ {(a-1)}^2& 2n^2& 4n-2\\ {(a-1)}^3& 3n^3& 3n^2-3n\end{array}\right|$ . Then ${\sum }_{a-1}^n△a$ is equal to

• A. $0$
• B. $(a-1)\sum n^2$
• C. $(a-1)n\sum n$
• D. None of these

Answer 1

Answer: (A)

$0$

$\Delta a=\left|\begin{array}{ccc}a-1& n& 6\\ (a-1{)}^2& 2n^2& 4n-2\\ (a-1{)}^3& 3n^3& 3n^2-3n\end{array}\right|$

${\sum }_{a-1}^n\Delta a=\left|\begin{array}{ccc}n& n& 6\\ 2n^2& 2n^2& 4n-2\\ 3n^3& 3n^3& 3n^2-3n\end{array}\right|$

$C_1=C_2$

$\therefore {\sum }_{a-1}^n\Delta a=0$