​$$\text{Find the value of } \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{2}\right)^{-2}.$$​​

Correct option is D

Given:

​$$\left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{2}\right)^{-2}.$$

Formula Used:

aⁿ= $$\left(\frac 1a \right)^{-n}$$

Solution:

​$$\left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{2}\right)^{-2}\\ \ \\= 4^2+6^2+2^2 \ \\\\ = 16 +36 +4 =56$$​​