Question

If $f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x\left(x-1\right)& x\left(x+1\right)\\ 3x\left(x-1\right)& x\left(x-1\right)\left(x-2\right)& x\left(x-1\right)\left(x+1\right)\end{array}\right|$ then $f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(100\right)$ is equal to

• A. $5$
• B. $8$
• C. $0$
• D. $10$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

Finding the value of $f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(100\right)$:

Given that,

$f\left(x\right)=\left|\begin{array}{ccc}1& x& x+1\\ 2x& x\left(x-1\right)& x\left(x+1\right)\\ 3x\left(x-1\right)& x\left(x-1\right)\left(x-2\right)& x\left(x-1\right)\left(x+1\right)\end{array}\right|$

By taking common $x$ and $x\left(x-1\right)$ from row $R_2$ and $R_3$,

$f\left(x\right)=x\left(x\left(x-1\right)\right)\left|\begin{array}{ccc}1& x& x+1\\ 2& \left(x-1\right)& \left(x+1\right)\\ 3& \left(x-2\right)& \left(x+1\right)\end{array}\right|$

Now, take common $\left(x+1\right)$ from the column $C_3$.

$\begin{array}{rcl}f\left(x\right)& =& x\left(x\left(x-1\right)\right)\left(x+1\right)\left|\begin{array}{ccc}1& x& 1\\ 2& \left(x-1\right)& 1\\ 3& \left(x-2\right)& 1\end{array}\right|\\ & =& x^2\left(x^2-1\right)\left|\begin{array}{ccc}2-1& x-1-x& 1-1\\ 3-2& x-2-x+1& 1-1\\ 3& \left(x-2\right)& 1\end{array}\right|\left[R_1\to R_2-R_1,R_2\to R_3-R_2\right]\\ & =& x^2\left(x^2-1\right)\left|\begin{array}{ccc}1& -1& 0\\ 1& -1& 0\\ 3& \left(x-2\right)& 1\end{array}\right|\end{array}$

If any two rows or columns of a determinant are identical, then the value of the determinant is $0$.

So, for any value of $x$,

$f\left(x\right)=0$, then

$f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(100\right)=0$

Hence, the correct option is C.