From a recurrence to an explicit term

$u_1=1$, $u_2=2(1)+1=3$, $u_3=2(3)+1=7$, $u_4=2(7)+1=15$, $u_5=2(15)+1=31$.

$u_2=2^1+2^0$, $u_3=2^2+2^1+2^0$, $u_4=2^3+2^2+2^1+2^0$, $u_5=2^4+2^3+2^2+2^1+2^0$. Thus

$\begin{aligned}&u_n\\ & =2^0+2^1+\cdots+2^{n-1}\\ & =\frac{1(2^n-1)}{2-1}\\ & =2^n-1.\end{aligned}$

In sequence mode set the initial index to 1, initial value $u_1=1$, and enter the recurrence as $u(n)=2u(n-1)+1$. The table gives 1,3,7,15,31; observing $1=2^1-1$, $3=2^2-1$ and so on suggests $u_n=2^n-1$. The expanded geometric sum above establishes the formula; a finite calculator table alone is evidence of a pattern, not a proof.

The calculator screenshot continues the table through n=11:

The expanded first five terms above answer the requested part; the remaining rows retain the source calculator display.