Question:
Tick (✓) 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) $\frac{144}{61}$
(C) 29
(d) $\frac{1}{29}$
Solution:
(C) 29
$\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}=\left(\frac{2}{1}\right)^{2}+\left(\frac{3}{1}\right)^{2}+\left(\frac{4}{1}\right)^{2}$
$=2^{2}+3^{2}+4^{2}$
$=4+9+16$
$=29$
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