Simply Simple

Observe a pattern -

{ 2 }^{ 5 }-{ 2 }^{ 4 }=16={ 2 }^{ 4 }

and

{ 2 }^{ 3 }-{ 2 }^{ 2 }={ 2 }^{ 2 }

So for any

{ 2 }^{ n }-{ 2 }^{ n-1 }={ 2 }^{ n-1 }

Or you can think about it in this way -

2^{ 5 }-{ 2 }^{ 4 }=2\times { 2 }^{ 4 }-2^{ 4 }\\ { 2 }^{ 4 }(2-1)={ 2 }^{ 4 }

Therefore, if you simplify the above pattern, you can observe the same pattern -

${ 2 }^{ 2015 }-{ 2 }^{ 2014 }={ 2 }^{ 2014 }\\ { 2 }^{ 2014 }-{ 2 }^{ 2013 }= { 2 }^{ 2013 }...$

And this can be reduced to

${ 2 }^{ 2 }-{ 2 }^{ 1 }={ 2 }^{ 1 }\\ { 2 }^{ 1 }-{ 2 }^{ 0 }={ 2 }-1=1$

Using $\large 1+x +x^2 +...+x^n = \frac{x^{n+1} -1}{x-1}$ ... $2^{2015} - (2^{2014} + 2^{2013} + 2^{2012}+...+ 1) = 2^{2015} -(2^{2015} - 1) = \boxed{1}$

Many people are taking the forward approach to this problem (as did I), but I have an alternate solution. We will claim the answer is $1$ and prove this result.

Let $2^{2015}-2^{2014}-2^{2013}-\ldots-2^0=1.$ Rearranging, we find that $1+2^0+2^1+\ldots+2^{2013}+2^{2014}=2^{2015}.$ We will now inspect the left hand side.

The first two terms are $1$ and $2^0=1.$ Add these together to get $2=2^1.$ The first two terms of our new expression are $2^1$ and $2^1.$ Since $2^k+2^k=2\times2^k=2^{k+1},$ the first two terms of the newest sequence are $2^2$ and $2^2.$ We can repeat this over and over again until we reach $2^{2014}+2^{2014}.$ Since $2^{2014}+2^{2014}=2^{2015},$ the first term of our original sequence was in fact $\boxed{1}.$

If you're having trouble visualizing what I am describing, this may help.

$\begin{aligned} 2^{2015}&=\left(1+2^0\right)+2^1+2^2+\ldots+2^{2014}\\ &=\left(2^1+2^1\right)+2^2+2^3+\ldots+2^{2014}\\ &=\left(2^2+2^2\right)+2^3+2^4+\ldots+2^{2014}\\ &=\vdots\\ &=\left(2^{2013}+2^{2013}\right)+2^{2014}\\ &=2^{2014}+2^{2014}\\ 2^{2015}&=2^{2015} \end{aligned}$

The given expression =

2^2015 - [ 1+ 2 + 2^2 + 2^3 + ...... + 2^2014] =

2^2015 - [ sum of terms of a geometric sequence of common ratio 2 and number

of terms 2015] =

2^2015 - (2^2015 - 1)/(2 - 1) = 1

Let the sum be S, multiply both sides by 2 and then subtract 1st series from the 2nd to get 2S -S = 2^2016 - 2^2015 - 2^2015 + 2^0 = 2^0 = 1

Python:

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answer = 2**2015
n = 2014
while n >= 0:
    answer -= 2**n
    n -= 1
print "Answer:", answer

Use this formula 2^n - 2^n-1 = 2^n-1.

Notice that 2^2015 = 2^2016 - 2^2015 2^2014 = 2^2015-2^2014 2^0 = 2^1 - 2^0 So 2^2015 - 2^2014 -...-2^0 = (2^2016-2^2015) - (2^2015-2^2014)-(2^2014-2^2013) - (2^2 - 2^1 ) - (2^1 - 2^0) = 2^2016 - 2^2015 - 2^2015 + 2^0 As 2• 2^2015 = 2^2016 the answer is 2^0 = 1