Answer
$x_n=2^{n-2}$; for all $n \geq 2$
Work Step by Step
According to the Principle of Mathematical Induction $a_n \geq 0$. This implies that there exist $n$ for all natural numbers N.
Given: $x_{n+1}=x_1+x_2+...+x_n$ and $x_1=1$
and, $x_2=x_1=1$; $x_3=x_1+x_2=1+1=2$ and $x_4=x_1+x_2+x_3=1+1+2=4....$
In the same manner, we have $x_1=1,x_2=2^0,x_3=2^1....$
We can see that the pattern we follow for sequence $x_n$ will be: $x_n=2^{n-2}$; for all $n \geq 2$
