I understoof the logic behind it but have no idea how to put it into words.
ex: n=3 ther are 4 n sum series: 1,1,1 1,2 2,1 3 of natural numbers different then 0, will be called n sum series if the sum of all its variables equals n. you need to proof that for every n>0 there are exactly 2^(n-1) series .
Thank you for helping.
Consider the set $\{1,2,\ldots,n\}$. The set of all subsets of this set has $2^n$ elements. A subset is equivalent to a string of $n$ 0s and 1s where the $i$ th number is 0 if $i$ is not contained in the set and 1 if it is.
From a subset we can construct a sequence of numbers that sum to $n$ as follows. Say the string starts with a 0. There will be some number of consecutive 0s, and we let that number be the first term in our sum. Then we let the next number be the size of the next block, which will be of consecutive 1s. Continuing in this fashion we obtain a sequence of numbers whose length is the number of blocks and whose sum is n. In fact we well get all such sequences this way. However, we have double counted them all because we can flip 1 and 0 to get the same sum. Thus the number of sequences is $2^n/2=2^{n-1}$.
