Question

Mark (✓) against the correct answer

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=?$

(a) $\frac{61}{144}$
(b) 29
(c) $\frac{144}{61}$
(d) none of these

Answer 1

(b) 29

We have:

${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}={\left(\frac{2}{1}\right)}^2+{\left(\frac{3}{1}\right)}^2+{\left(\frac{4}{1}\right)}^2$ $\left[s\mathrm{ince}{\left(\frac{a}{b}\right)}^{-1}={\left(\frac{b}{a}\right)}^1\right]$
= (2² + 3² + 4²)
= (4 + 9 + 16)
= 29