TITLE: Prove that $\pm1\pm2\pm\ldots\pm(4n+1)$ yields all odd numbers up to $(2n+1)(4n+1)$
QUESTION [5 upvotes]: Problem: Prove that for different choices of signs $+$ and $-$ the expression $$\pm1\pm2\pm3\pm4\pm5...\pm(4n+1)$$ yields all odd positive integers less than or equal to $(2n+1)(4n+1).$
My Attempt: Let $n=1$, then we have the following expression to work with $$\pm1\pm2\pm3\pm4\pm5.$$
Let $+$ be represented by $0$ and $-$ be represented by $1.$ A binary string of length $5$ therefore represents the permutation of the operations. So $$+1+2+3+4+5\text{ is equivalent to }00000$$ 
$$+1-2+3+4+5\text{ is equivalent to }01000$$ and so on. With this representation we observe that:
$$15=00000$$
$$13=10000$$
$$11=01000$$
$$9=00100$$
$$7=00010$$
$$5=00001$$
$$3=10001$$
$$1=01001.$$ 
Notice how $1$ traverses in each number. Ignoring the trivial case when the odd number is equal to $(2n+1)(4n+1)$ (which in this case is $15$) we observe that there are $4*1+1$ numbers that can be written with only one $1$ since the numeral $1$ has $4*1+1$ places to move on. The remaining places $$\frac{1*(1+3)}{2}=2$$ require a two $1$s. 
This observation motivates the following Proof: Given any $n\geq 1$ we find $4n+1.$ Then we have $4n^2+3n$ odd numbers that are strictly less than $(2n+1)(4n+1).$ Let $a_i$ denote these odd numbers where $1\leq i\leq 4n^2+3n$ and $a_1<a_2<....<a_{4n^2+3n}.$ We begin by covering the last $4n+1$ odd numbers by writing 
$$a_{4n^2+3n}=\underbrace{1000000...0}_{4n+1}=8n^2+6n-1$$
$$..$$
$$a_{4n^2+3n-(4n)}=00000...1=8n^2+2n+1$$
$$a_{4n^2+3n-(4n+1)}=\underbrace{10000...1}_{4n+1}=8n^2+2n-1$$
$$..$$
and so on. Essentially $4n^2+3n=(4n+1)+(4n)+(4n-1)+(4n-2)+...+(4n-(n-1))+\underbrace{\frac{n(3+n)}{2}}_{\text{need }n+1\text{ 1s}}=4n^2+3n.$
Now I don't know how to compile these observations into a formal proof. What proof technique would be best suited in this scenario, maybe induction? Moreover, should I inclucde Lemmas regarding the zero and one arrangments?

REPLY [3 votes]: Note that $1+2+3+...+(4n+1)=(2n+1)(4n+1)$ $(1)$ and that is a odd number. So if we sum $-2$ at both sides we get:
$$-1+2+3+4+5+...+(4n+1)= (2n+1)(4n+1)-2$$
If we sum $-4$ at both sides of $(1)$ we get:
$$1-2+3+4+5+...+(4n+1)= (2n+1)(4n+1)-4$$
If we sum $-6$ at both sides of $(1)$ we get:
$$1+2-3+4+5+...+(4n+1)= (2n+1)(4n+1)-6$$
And we just follow that algorithm until we have to sum $-2(4n+1)$ and get
$$1+2+3+4+5+...-(4n+1)= (2n+1)(4n+1)-2(4n+1)$$
Now we have to sum $-2-2(4n+1)$ in order to get the next odd number:
$$-1+2+3+4+5+...-(4n+1)= (2n+1)(4n+1)-2(4n+1)-2$$
We can see now what is the algorithm that we have to follow in order to get all odd numbers from $1$ to $(2n+1)(4n+1)$.