Originally Posted by

**BobBali** Hi All & Good-day,

The problem states, Show that:

$\displaystyle (\frac{2}{x} + 4)^2 = \frac{1 + 2^x+3 + 2^2x+4}{2^2x }$

* Above Denimonator should be 2 power 2x (latex trouble)

$\displaystyle (2^{-x + 4})(2^{-x + 4}) $

$\displaystyle 2^{-2x + 4}(2^{-x}) + 4(2^{-x}) + 16$

$\displaystyle 2^{-2x} + 2^{-x} (4+4) + 16$

$\displaystyle 2^{-2x} + 2^{-x} (2^2 + 2^2) + 2^4$

$\displaystyle 2^{-2x} + 2^{-x+2} + 2^{-x+2} + 2^4$

After this i am at a loss... But i do know that if i take the above first term's

reciprocal > $\displaystyle \frac{1}{2^2x }$

I now have the denominator in the answer, but how to distinguish the 2nd and 3rd terms in the answer...?