## University Calculus: Early Transcendentals (3rd Edition)

$x_n=2^{n-2}$; for all $n \geq 2$
Since, we are given that $x_{n+1}=x_1+x_2+...+x_n$ and $x_1=1$ Then we get, $x_2=x_1=1$; $x_3=1+1=2$ and $x_4=1+1+2=4$ Likewise, we get $x_1=1,x_2=2^0,x_3=2^1$....so on Now, we have the formula for the sequence $x_n$ is $x_n=2^{n-2}$; for all $n \geq 2$