jabhatta2
lpetroski
If n is a positive integer, then \((-2^{n})^{-2}\) +\((2^{-n})^{2}\) =
A. 0
B. \(2^{(-2n)}\)
C. \(2^{(2n)}\)
D. \(2^{(-2n+1)}\)
E. \(2^{(2n+1)}\)
Hi
Bunuel zhanbo -- Wanted to focus on the first expression (marked in yellow)
When I inserted [n=1] into the yellow function : I got (
-1/4)
But that is clearly wrong as I should be getting [1 / 4 ] if i insert n = 1 into the yellow expression.
Here is how I got ( - 1 / 4)
Could you please let me know, where is the mistake in my step by step ?
The point is that \((-2^{n})^{-2}\) means \((-1*2^{n})^{-2}\). So, the minus sign there just means "multiplied by -1" and the exponent of -2 should be applied to -1 too. Thus, if we go the way you are doing we'd get:
\((-2^{n})^{-2} =(-1*2^{n})^{-2}=(-1)^{-2}*2^{-2n}= 1*2^{-2n}=2^{-2n}=\frac{1}{2^{2n}} \)
Now, if you plug 1 for n, you'd' get 1/4.
Here is how I'd solve this question.
If n is a positive integer, then \((-2^{n})^{-2}+(2^{-n})^{2} = \)A. 0
B. \(2^{(-2n)}\)
C. \(2^{(2n)}\)
D. \(2^{(-2n+1)}\)
E. \(2^{(2n+1)}\)
Because -2 is even, then \((-2^{n})^{-2}\) is the same as \((2^{n})^{-2}\). So, we have:
\((-2^{n})^{-2}+(2^{-n})^{2} = \)
\(=(2^{n})^{-2}+(2^{-n})^{2} =\)
\(=2^{(-2n)}+2^{(-2n)} =\)
\(=2*2^{(-2n)} =\)
\(=2^{(-2n+1)}\).
Answer: D.
Hope it helps.