The first term of a sequence is $a_{1}>1$. Any subsequent term of the sequence can be obtained by subtracting one from twice the term that immediately precedes it. Write a formula for the sum of the first $n$ terms.
According to the rule for generating the terms of the sequence, $a_{2}=2 a_{1}-1$. It is useful to write this as $a_{2}=2\left(a_{1}-1\right)+1$. What do we gain by this? If we generate $a_{3}$ from this form, it immediately follows that: $a_{3}=4\left(a_{1}-1\right)+1$. (Twice 1 minus 1 again results in 1.) It is also easy to see that for every positive integer $k$, $a_{k}=2^{k-1}\left(a_{1}-1\right)+1$. With this formula, we can write down any term of the sequence without having to calculate the previous terms.

Let's turn to the question of the problem. Let's write down the sum of the first $n$ terms:

$$
S_{n}=a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}
$$

Multiply both sides of the equality by 2, and subtract $n$ from both sides (on the right side, $n$ ones).

$$
2 S_{n}-n=\left(2 a_{1}-1\right)+\left(2 a_{2}-1\right)+\cdots+\left(2 a_{n-1}-1\right)+\left(2 a_{n}-1\right) .
$$

According to the rule for generating the sequence,

$$
2 S_{n}-n=a_{2}+a_{3}+\cdots+a_{n}+a_{n+1}
$$

Subtract $S_{n}$ from both sides:

$$
S_{n}-n=a_{n+1}-a_{1}
$$

Using our formula for $a_{k}$: $a_{n+1}=2^{n}\left(a_{1}-1\right)+1$, therefore

$$
S_{n}=\left(a_{1}-1\right)\left(2^{n}-1\right)+n
$$