Finding a formula for the sums $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n$

Question

I am told to consider the sum $2^0 + 2^1 + 2^2 + 2^3 + \cdots + 2^n$ for $n ∈ ℕ ∪ \{0\}$. Next, I am to find the value of the sum for $n= 0, 1, 2, 3, 4$. After I find those values, I am supposed to come up with a conjecture that gives a simple closed formula. When I did the sums, I realize that the difference between sums doubles every time. So,

$n=0$ is $1$,

$n=1$ is $3$,

$n=2$ is $7$,

$n=3$ is $15$,

$n=4$ is $31$.

My problem is that I am having a hard time trying to come up with a conjecture. How would I be able to come up with one? Is it just Guess and Check?

Answer 1

As the difference doubles every (they are 2, 4, 8, 16) we have $x_{n+1}=x_n+2^{n+1}$. So using this we can see $$x_n=\sum_{n=0}^n2^{n}$$

This is a finite geometric series, so it is given by \begin{align*}x_n=\sum_{n=0}^n2^{n}&=\frac{1-2^{n+1}}{1-2} \\&=2^{n+1}-1\end{align*}

Answer 2

Hint

In a more general manner, if $$x_n=\sum_{n=0}^n k^{n}$$ you face a geometric progression and whatever $k$ could be (except $k=1$) you should end with $$x_n=\sum_{n=0}^n k^{n}=\frac{k^{n+1}-1}{k-1}$$

Still more general, if $$x_n=\sum_{m}^n k^{n}=\frac{k^{n+1}-k^m}{k-1}$$

Answer 3

Without words:

$$2S=2(2^0+2^1+\cdots 2^n)=2^1+2^2+\cdots 2^{n+1}=-2^0+S+2^{n+1}.$$

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