Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:
$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$
Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:
$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$
Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.
Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$
If it is the case you want the sequence to start with $a_1$ we need to shift this result to the right one:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(1) {n-1 \choose i}$$
Note when you re-define $a_0$ to be $a_1$ by shifting it's index to the right one, you're re-defining $a_1-a_0=\Delta^1(0)$ to be $a_2-a_1=a_{1+1}-a_{1}=\Delta^1(1)$.
Best Answer
$$U_2-U_1=2^2$$
$$U_3-U_2=2^3$$
$$U_4-U_3=2^4$$
So by proceeding in this manner we get that, $$U_n-U_{n-1}=2^n$$
Now by adding these equations we get
$U_n-U_1= 2^2+2^3+2^4+.....+2^{n-1}+2^n \Rightarrow U_n=U_1+2^2+2^3+2^4+.....+2^{n-1}+2^n$
$U_n=2^1+2^2+2^3+2^4+.....+2^{n-1}+2^n$
Therefore $$U_n=\sum_{k=1}^{n}2^k$$
$$U_n=\frac{2^{n+1}-2}{2-1}$$
$$U_n=2^{n+1}-2$$