Question

If ${\Delta }_r=\left|\begin{array}{ccc}{2}^n+1& 2^{n+1}& n(n+2)\\ 1& 1& -1\\ {}^n{C}_r& 2^r& 2r+1\end{array}\right|$
then the value of $\sum _{r=0}^n{\Delta }_r$ is

• A. $2$
• B. $0$
• C. a function of $n$
• D. $1$

Answer 1

The correct option is B $0$

${\Delta }_r=\left|\begin{array}{ccc}{2}^n+1& 2^{n+1}& n(n+2)\\ 1& 1& -1\\ {}^n{C}_r& 2^r& 2r+1\end{array}\right|$

${\sum }_{r=0}^n{\Delta }_r=\left|\begin{array}{ccc}{2}^n+1& 2^{n+1}& n(n+2)\\ 1& 1& -1\\ {\sum }_{r=0}^n{}^n{C}_r& {\sum }_{r=0}^n{2}^r& {\sum }_{r=0}^n(2r+1)\end{array}\right|$

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

Applying $C_2\to C_2-C_1,C_3\to C_3+C_1$

${\sum }_{r=0}^n{\Delta }_r=\left|\begin{array}{ccc}{2}^n+1& 2^{n+1}-2^n-1& n(n+2)+2^n+1\\ 1& 0& 0\\ 2^n& 2^{n+1}-2^n-1& (n+1)(n-1)+2^n\end{array}\right|$

On expanding above determinant we get,

$=0$