Find a sequence

Computing values for $n=2,3,4$ we see

$x_2=2^2-1$

$x_3=2^4-1$

$x_4=2^8-1$

So the sequence can be shown by induction to be $\displaystyle\,x_n=2^{2^{n-1}}-1$ . So plugging in $n=7$ we find $\boxed{x_7=2^{64}-1}$ .

I'll add a small but very useful observation that hasn't yet been mentioned. We can complete the square in the given iteration formula to bring it in this form: $x_{n+1} = x_{n}^{2} + 2x_{n} = (x_{n}^{2}+2x_{n} +1) - 1 = (x_{n} + 1)^2 - 1 \Leftrightarrow \boxed{x_{n+1}=(x_{n}+1)^2 -1}$

Now we can very easily plug in values for $x_{1},x_{2},...,x_{6}$ . Here are the steps in detail:

$x_{2}=(1+1)^{2}-1=2^{2}-1$ Do NOT make the mistake of writing this as $x_{2}=5$ . We are working with powers of $2$ here.
$x_{3}=(2^{2}-1+1)^{2}-1=2^{4}-1$
$x_{4}=(2^{4}-1+1)^{2}-1=2^{8}-1$
$x_{5}=(2^{8}-1+1)^{2}-1=2^{16}-1$
$x_{6}=(2^{16}-1+1)^{2}-1=2^{32}-1$
$x_{7}=(2^{32}-1+1)^{2}-1=2^{64}-1 \Leftrightarrow \boxed{x_{7}=2^{64}-1}$