DON'T USE A CALCULATOR !!!!!

I just grabbed a pen and paper and got stuck in. It was actually a bit of fun that way.

a = 2, r = 4/2 = 2

$S_{n}=\frac{a(r^{n}-1)}{r-1} S_{13}=\frac{2(2^{13}-1)}{2-1}$ = 2 x (8392-1)
= 2 x 8391 = $\boxed{16382}$

*I 'm sorry for bad formatting. Maybe someone can help me to tidy up my formatting.

Simply solved as this ia an G.P. and we can simply use the expression of sum of GP.......... Yeah. :)

We will prove that $\sum _{ k=0 }^{ n }{ { 2 }^{ k } } ={ 2 }^{ k+1 }-1$ using induction.

Base Case;

When $k=0$ , $\sum _{ k=0 }^{ n }{ { 2 }^{ k } } = 2^0 = 1 = 2-1 = 2^1-1$

Inductive Step

Assume that $\sum _{ k=0 }^{ n }{ { 2 }^{ k } } ={ 2 }^{ k+1 }-1$ for some $k \in \mathbb{Z}$

We will prove that (\sum _{ k=0 }^{ n }{ { 2 }^{ k+1 } } ={ 2 }^{ k+2 }-1).

Multiplying the assumption by 2,

$2 \cdot 2^0+2 \cdot 2^1+...+2 \cdot 2^{k}={ 2 }^{ k+2 }-2$

Adding $1$ to both sides,

$1+2^1+2^2+...+2^{k+1}={ 2 }^{ k+2 }-1$

So $\sum _{ k=0 }^{ n }{ { 2 }^{ k+1 } } ={ 2 }^{ k+2 }-1$ .

So we have the conclusion, and the answer is $2^{14}-1=\boxed{16383}$ $\square$

2^13=2^14-2=16382

in PHP:

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$sum = 0;
        for ($i = 1; $i <= 13; $i++) {
            $sum += pow(2,$i);
        }
        print_r($sum);
        exit;

     // prints 16382

..but if we would utilize the values in the table given, we could see a pattern and come up with a formula for the summation of 2^n.

X = (2^(n+1)) - 2

where n = 13

X = (2^(13+1))-2

X = (2^14)-2

X = 16384-2

X = 16382

Stop writing in all caps. It hurts my eyes.

mere observation !

just subtract two from value 2^14

I honestly just added them all. EDIT:I have actually found a way to do this!!! $2^k-1=(2-1)(2^{k-1}+2^{k-2}+\dots+2^1+1)\\ 2^k-1=2^{k-1}+2^{k-2}+\dots+1$ Insert k=14.The R.H.S becomes: $2^{13}+2^{12}+\dots+1$ In order for this to become the series we want, we have to subtract 1.So: $2^{14}-2=2^{13}+2^{12}+\dots+2^1=16384-2=\boxed{16382}$