Question

If ${\Delta }_r=\left|\begin{array}{ccc}{a}^r& b^r& c^r\\ a& b& c\\ 1-a& 1-b& 1-c\end{array}\right|$. If $\sum _{r=0}^{\infty }{\Delta }_r=0$, then which statement is true.

• A. $a=b=c,a\in R$
• B. $a=b=c,\left|a\right|<1$
• C. $a=b=c\ne 1$
• D. $a=b\ne c,\left|c\right|<1$

Answer 1

The correct option is C $a=b=c\ne 1$

${△}_r=\left|\begin{array}{ccc}{a}^r& b^r& c^r\\ a& b& c\\ 1-a& 1-b& 1-c\end{array}\right|$

${\sum }_{r=0}^{\infty }{△}_r=0$

$\Rightarrow \left|\begin{array}{ccc}{\sum }_{r=0}^{\infty }{a}^r& {\sum }_{r=0}^{\infty }{b}^r& {\sum }_{r=0}^{\infty }{c}^r\\ a& b& c\\ 1-a& 1-b& 1-c\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}\frac{a}{1-a}& \frac{b}{1-b}& \frac{c}{1-c}\\ a& b& c\\ 1-a& 1-b& 1-c\end{array}\right|=0$

by $R_3\to R_3+R_2$

$\Rightarrow \left|\begin{array}{ccc}\frac{a}{1-a}& \frac{b}{1-b}& \frac{c}{1-c}\\ a& b& c\\ 1& 1& 1\end{array}\right|=0$

by $c_3\to c_3-c_1\&c_2\to c_2-c_1$

$\Rightarrow \left|\begin{array}{ccc}\frac{a}{1-a}& \frac{b}{1-b}& \frac{c}{1-c}\\ a& b-a& c-a\\ 1& 0& 0\end{array}\right|=0$

$\Rightarrow (\frac{b}{1-b}-\frac{a}{1-a})(c-a)-\{\frac{c}{1-c}-\frac{a}{1-a}\}(b-a)=0$

$\Rightarrow \left\{\frac{b-ab-a+ab}{(1-b)(1-a)}\right\}(c-a)-\left\{\frac{c-ca-a+ca}{(1-c)(1-a)}\right\}(b-a)=0$

$\Rightarrow \frac{(b-a)(c-a)}{(1-b)(1-a)}-\frac{(c-a)(b-a)}{(1-c)(1-a)}=0$

$\Rightarrow \frac{(b-a)(c-a)}{(1-a)}\{\frac{1}{1-b}-\frac{1}{1-c}\}=0$

$\Rightarrow \frac{(b-a)(c-a)}{(1-a)}\left\{\frac{1-c-1+b}{(1-b)(1-c)}\right\}=0$

$\Rightarrow \frac{(b-a)(c-a)(b-c)}{(1-a)(1-b)(1-c)}=0$

so, $a=b=c\ne 1.$

Hence, the answer is $a=b=c\ne 1.$