induction proof: number of compositions of a positive number $n$ is $2^{n-1}$

Question

So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction. $$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$ When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$ $$ 1+2+4+(2^{3}+1)=2^{4}$$ $$ 1+2+4+(8+1)=16$$ $$ 16=16$$ Assume that $n=k$ $$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$ Now show for $k+1$ $$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$ Don't know what to do after. Help is appreciated. $$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$

Answer 1

You actually don't need induction to prove this. There's a telescoping effect.

$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$

Carry the one.

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

$1 = 2^0$

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

Collapse the parentheses

$$(2^1 + 2^1) + 2^2 + ... + 2^{n-2}=2^{n-1}$$

Collapse the parentheses.

$$(2^2 + 2^2) + ... + 2^{n-2}=2^{n-1}$$

Keep doing this until

$$2^{n-2} + 2^{n-2} = 2^{n-1}$$.

Answer 2

Let's return to the original problem. A composition of a positive integer $n$ is a way of writing $n$ as the sum of a sequence of strictly positive integers.

For $n = 5$, there are indeed $2^{5 - 1} = 2^4 = 16$ compositions since each composition corresponds to placing or omitting an addition sign in the four spaces between successive ones in a row of five ones.

\begin{array}{c c} \text{composition} & \text{representation}\\ \hline 5 & 1 1 1 1 1\\ 4 + 1 & 1 1 1 1 + 1\\ 3 + 2 & 1 1 1 + 1 1\\ 3 + 1 + 1 & 1 1 1 + 1 + 1\\ 2 + 3 & 1 1 + 1 1 1\\ 2 + 2 + 1 & 1 1 + 1 1 + 1\\ 2 + 1 + 2 & 1 1 + 1 + 1 1\\ 2 + 1 + 1 + 1 & 1 1 + 1 + 1 + 1\\ 1 + 4 & 1 + 1 1 1 1\\ 1 + 3 + 1 & 1 + 1 1 1 + 1\\ 1 + 2 + 2 & 1 + 1 1 + 1 1\\ 1 + 2 + 1 + 1 & 1 + 1 1 + 1 + 1\\ 1 + 1 + 3 & 1 + 1 + 1 1 1\\ 1 + 1 + 2 + 1 & 1 + 1 + 1 1 + 1\\ 1 + 1 + 1 + 2 & 1 + 1 + 1 + 1 1\\ 1 + 1 + 1 + 1 + 1 & 1 + 1 + 1 + 1 + 1 \end{array}

Using this observation, we can write a combinatorial proof. A positive integer $n$ has $2^{n - 1}$ compositions since each composition is uniquely determined by choosing to include or omit an addition sign in each of the $n - 1$ spaces between successive ones in a row of $n$ ones.

Answer 3

$n=k$ $$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$ so

$$\underbrace{2^0+2^1+2^2+...+2^{k-2}}=\overbrace{2^{k-1} -1}$$

Now show for k+1 $$\underbrace{2^0+2^1+2^2+...+2^{k-2}}+(2^{k-1}+1)=\overbrace{2^{k-1} -1}+(2^{k-1}+1)$$

$$=2^{k-1}+2^{k-1}=2*2^{k-1}=2^{k}$$

Answer 4

You're almost there. You just have to use your hypothesis in your final case for $k+1$.

You found for $n=k$ that $$2^0+2^1+2^2+...+2^{k-2}+1=2^{k-1}.$$ You want to prove for $n=k+1$ that $$2^0+2^1+2^2+...2^{k-2}+2^{k-1}+1=2^{k},$$ Fill in the first equation into the second \begin{align*} 2^0+2^1+2^2+...+ 2^{k-2}+&(2^0+2^1+2^2+...+2^{k-2}+1)+1\\ =2&(2^0+2^1+2^2+...+2^{k-2}+1) \end{align*} Now using the induction hypothesis again \begin{align*} &=2(2^{k-1})\\&=2^{k}. \end{align*}

By mathematical induction we have now proven this for all $n\geq 5$.

The theorem does however hold for any $n\geq 2$, so I suggest making your basecase be $n=2$.

Answer 5

What you know: $\color{blue}{2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}}$

What you want to show: $\color{green}{2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}}$

Use what you know:

$\color{green}{2^0+2^1+2^2+...+2^{k-2} + (2^{k-1}+1)} =$

$\color{blue}{2^0+2^1+2^2+...+2^{k-2}} + (\color{green}{2^{k-1}} + \color{blue}1) =$

$\color{blue}{2^0+2^1+2^2+...+(2^{k-2}+1)}+\color{green}{2^{k-1}}=$

$\color{blue}{2^{k-1}} + \color{green}{2^{k-1}}=$

$\color{green}{2\times 2^{k-1}} =$

$\color{green}{2^{(k-1) + 1}} =$

$\color{green}{2^k}$.