Question

If $a=(2^{-2}-2^{-3}),b=(2^{-3}-2^{-4})andc=(2^{-4}-2^{-2})$ then find the value of 3 abc

• A. $\frac{-63}{1024}$
• B. $\frac{-63}{2048}$
• C. $\frac{-9}{2048}$
• D. $\frac{9}{1024}$

Answer 1

The correct option is C $\frac{-9}{2048}$

$a=(2^{-2}-2^{-3}),b=(2^{-3}-2^{-4})andc=(2^{-4}-2^{-2})$
So, $3abc=3\times (2^{-2}-2^{-3})\times (2^{-3}-2^{-4})\times (2^{-4}-2^{-2})$
$3abc=3\times (1/4-1/8)\times (1/8-1/16)\times (1/16-1/4)$
$3abc=3\times \left(1/8\right)\times \left(1/16\right)\times (-3/16)$
$3abc=\frac{-9}{2048}$
Answer (C) $\frac{-9}{2048}$