Question

If $a=(2^{-2}-2^{-3}),b=(2^{-3}-2^{-4})$ and $c=(2^{-4}-2^{-2})$ then find :
(i) $a^3+b^3+c^3$ (ii) $10abc$

• A. $\frac{9}{2048},\frac{7}{2048}$
• B. $\frac{3}{1024},\frac{5}{2048}$
• C. $\frac{-3}{1024},\frac{-10}{2048}$
• D. $\frac{-9}{2048},\frac{-15}{1024}$

Answer 1

The correct option is D $\frac{-9}{2048},\frac{-15}{1024}$

$a=(2^{-2}-2^{-3});b=(2^{-3}-2^{-4});c=(2^{-4}-2^{-2})$

$a=\frac{1}{2^2}-\frac{1}{2^3}=\frac{2-1}{2^3}=\frac{1}{2^3}$

$b=\frac{1}{2^3}-\frac{1}{2^4}=\frac{2-1}{2^4}=\frac{1}{2^4}$

$c=\frac{1}{2^4}-\frac{1}{2^2}=\frac{1-4}{2^4}=\frac{-3}{2^4}$

i) $a^3+b^3+c^3=\frac{1}{(2^3{)}^3}+\frac{1}{(2^4{)}^3}+{\left(\frac{-3}{2^4}\right)}^3$

$=\frac{1}{2^9}+\frac{1}{2^{12}}-\frac{27}{2^{12}}$

$=\frac{2^3+1-27}{2^{12}}=\frac{-18}{2^{12}}=-\frac{9}{2048}$

$=-\frac{9}{2048}$

ii) $10abc=10\times \frac{1}{2^3}\times \frac{1}{2^4}\times \frac{-3}{2^4}$

$=\frac{-30}{2^{11}}=\frac{-15}{1024}$