Evaluate $$\left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{5}\right)^{-2}$$​​

Correct option is D

Let's evaluate the expression step by step:

(12)−2+(13)−2+(14)−2+(15)−2\left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{5}\right)^{-2}(21​)−2+(31​)−2+(41​)−2+(51​)−2​$$a^{-n} = \frac{1}{a^n}, so:$$​​

Recall that

(12)−2=22=4\left(\frac{1}{2}\right)^{-2} = 2^2 = 4(21​)−2=22=4​

(13)−2=32=9\left(\frac{1}{3}\right)^{-2} = 3^2 = 9(31​)−2=32=9​

(14)−2=42=16\left(\frac{1}{4}\right)^{-2} = 4^2 = 16(41​)−2=42=16​

(15)−2=52=25\left(\frac{1}{5}\right)^{-2} = 5^2 = 25(51​)−2=52=25​

Now sum them up:

​4 + 9 + 16 + 25 = 54​

So, the answer is 54, which corresponds to option (D).