We have
$$ n \in R $$
And an arithmetic sequence of natural numbers different than 0 that the sum of all its members = n
$$
a_1,a_2,...,a_k
$$
$$
a_1+a_2+...+a_k = n
$$
I need to prove by induction that for each n>0 there will be only
$$2^{n-1}$$
groups like this.
For example:
For n=3 we have only 4 groups like this:
1,1,1
1,2
2,1
3
How do I prove this by induction??
Thanks
Hint (also see my comment):
Here $\mathbb N=\{1,2,\dots\}$
Tuple $\left\langle a_{1},\cdots,a_{k}\right\rangle \in\mathbb{N}^{k}$ with $a_{1}+a_{2}+\cdots+a_{k}=n$ induces the tuples $\left\langle 1,a_{1},\cdots,a_{k}\right\rangle \in\mathbb{N}^{k+1}$ and $\left\langle a_{1}+1,\cdots,a_{k}\right\rangle \in\mathbb{N}^{k}$.
This with $1+a_{1}+\cdots+a_{k}=n+1=\left(a_{1}+1\right)+\cdots+a_{k}$.
So stepping from $n$ to $n+1$ doubles the number of tuples.
edit:
Let $S_{n,k}:=\left\{ \left\langle a_{1},\cdots,a_{k}\right\rangle \in\mathbb{N}^{k}\mid a_{1}+\cdots+a_{k}=n\right\} $ and $S_{n}=\bigcup_{k=1}^{n}S_{n,k}$ .
To be shown (by induction) is that $m_{n}=2^{n-1}$ where $m_{n}$ denotes the cardinality of $S_{n}$.
It is evident that $m_{1}=1=2^{0}$.
From $m_{n}=2^{n-1}$ (induction hypothese) it will follow immediately that $m_{n+1}=2^{n}$ if we can prove that $m_{n+1}=2m_{n}$. So proving the last statement will be enough.
Prescribe function $f:S_{n+1}\rightarrow S_{n}$ by $\left\langle a_{1},a_{2},\cdots,a_{k}\right\rangle \mapsto\left\langle a_{1}-1,a_{2},\cdots,a_{k}\right\rangle $ if $a_{1}>1$ and $\left\langle 1,a_{2},\cdots,a_{k}\right\rangle \mapsto\left\langle a_{2},\cdots,a_{k}\right\rangle $ otherwise.
Then $f$ is a surjective function with $f^{-1}\left(\left\langle b_{1},\cdots,b_{k}\right\rangle \right)=\left\{ \left\langle 1,b_{1},\cdots,b_{k}\right\rangle ,\left\langle b_{1}+1,\cdots,b_{k}\right\rangle \right\} $ for every element $\left\langle b_{1},\cdots,b_{k}\right\rangle \in S_{n}$.
Then $\left\{ f^{-1}\left(\left\langle b_{1},\cdots,b_{k}\right\rangle \right)\mid\left\langle b_{1},\cdots,b_{k}\right\rangle \in S_{n}\right\} $ forms a partition of $S_{n+1}$. It contains $m_{n}$ sets and each of these sets contains $2$ elements of $S_{n+1}$. This tells us that $m_{n+1}=2m_{n}$.