Question

If $x=2+5i($where $1i=\sqrt{-1})$ and $2(\frac{1}{1!9!}+\frac{1}{3!7!})+\frac{1}{5!5!}=\frac{2^a}{b!}$ then $x^3-5x^2+33x-10=$

• A. $a+b$
• B. $b-a$
• C. $a-b$
• D. $-a-b$
• E. $(a-b)(a+b)$

Answer 1

The correct option is A $a+b$

$\frac{10!2^a}{b!}=2{[}^{10}{C}_1{+}^{10}{C}_3]{+}^{10}{C}_5$
$\frac{10!2^a}{b!}{=}^{10}{C}_1{+}^{10}{C}_3{+}^{10}{C}_5{+}^{10}{C}_7{+}^{10}{C}_9=2^9$
$\therefore a=9,b=10$
$x=2+5i$
$(x-2{)}^2=-25$
$x^2-4x+29=0$
$\therefore x^3-5x^2+33x-10=(x-1)(x^2-4x+29)+19$
$\Rightarrow 19=a+b$