Proving the upper bound of the 2-power series

Question

I want to show that the sum of integers $2^1$, $2^2$, $2^3$ ... $2^n$ is bounded by 4N*, where N* is the value of n, the highest power used.

I've come at this by noting that $2^{n-2}$ + $2^{n-1}$ < $2^n$, and then that ($2^{n-5}$ + $2^{n-4}$) + ($2^{n-3}$ + $2^{n-2}$) < ($2^{n-1}$ + $2^{n}$), etc., thus reducing the numbers of terms in a sequence.

For example, we can see that, for N* = $2^{22}$,

($2^1$ + $2^2$ ... $2^{16}$) + ($2^{17}$ ... $2^{20}$)+ $2^{21}$+$2^{22}$ < 4($2^{22}$), with each term growing successively larger.

I can't help but shake the feeling, though, there is a more precise and rigorous way to demonstrate this relationship. Thoughts?

Answer 1

Because of $$2+2^2+2^3+...+2^n=2^{n+1}-2$$, the sum is even bounded by twice the highest power used. To see this equation, consider

$$S=2+2^2+2^3+...+2^n$$

$$2S=2^2+2^3+2^4+...+2^{n+1}$$

Now subtract those equations.

Answer 2

$$ \sum_{i=1}^n 2^i = 2\frac{1-{2^n}}{1-2} = 2^{n+1}-2 <2N*$$ by the formula for the sum of a geometric series.