Prove that $\vec{u_1}+\vec{u_2},\vec{u_2}+\vec{u_3},...,\vec{u_n}+\vec{u_1}$ is a basis of the vector space $V$, iff $n$ is odd Question's Claim:

Let $V$ be a vector space over a field $\mathbb{F}$, given that $\vec{u_1},\vec{u_2},...,\vec{u_n}$ are a base of $V$.
Prove that $\vec{u_1}+\vec{u_2},\vec{u_2}+\vec{u_3},...,\vec{u_{n-2}}+\vec{u_{n-1}},\vec{u_n}+\vec{u_1}$ are a base of the vector space $V$, iff $n$ is odd.


*

*Hint/Instruction: examine cases in which $n$ is a small number, e.g: $n=3,n=2$, and understand the difference between $n$ is odd, and even.


My attempt:
$Proof.$
$\implies:$ For the first direction, suppose that:$$\vec{u_1}+\vec{u_2},\vec{u_2}+\vec{u_3},...,\vec{u_{n-2}}+\vec{u_{n-1}},\vec{u_n}+\vec{u_1}$$ are a base of the vector space $V$. Assume towards a contradiction that $n$ is even, and thus, we get that for the multiplication by the scalar $-1\in \mathbb{R}$ of the vectors at the even location, and the multiplication by the scalar $1\in \mathbb{R}$ of the vectors at the odd location: $$\overrightarrow{u_{1}} +\overrightarrow{u_{2}} -\left(\overrightarrow{u_{2}} +\overrightarrow{u_{3}}\right) +\left(\overrightarrow{u_{3}} +\overrightarrow{u_{4}}\right) -\dotsc -\left(\overrightarrow{u_{n}} +\overrightarrow{u_{1}}\right)\\
\\
=\overrightarrow{u_{1}} +\overrightarrow{u_{2}} -\overrightarrow{u_{2}} -\overrightarrow{u_{3}} +\overrightarrow{u_{3}} +\overrightarrow{u_{4}} -\dotsc -\overrightarrow{u_{n}} -\overrightarrow{u_{1}}\\
\\
=\vec{0}$$
Therefore, we get a contradiction of the linearly independent, so when $n$ is even the vectors cannot be a base to $V$, and therefore $n$ is odd.
$\Longleftarrow:$ for the inverse direction, we suppose that $n$ is odd. So for the scalars $\beta_1,\beta_2,...,\beta_n \in \mathbb{F}$, we get:
$$\beta _{1}\left(\overrightarrow{u_{1}} +\overrightarrow{u_{2}}\right) +\beta _{2}\left(\overrightarrow{u_{2}} +\overrightarrow{u_{3}}\right) +\dotsc +\beta _{n}\left(\overrightarrow{u_{n}} +\overrightarrow{u_{1}}\right)\\
\\
=( \beta _{1} +\beta _{n})\overrightarrow{u_{1}} +\ ( \beta _{1} +\beta _{2})\overrightarrow{u_{2}} +( \beta _{2} +\beta _{3})\overrightarrow{u_{3}} +\dotsc +( \beta _{n} +\beta _{n-1})\overrightarrow{u_{n}}$$
By the given information, those scalars over the field $\mathbb{F}$ must be zero, since $\vec{u_1},\vec{u_2},...,\vec{u_n}$ are a base of $V$, so we have linear independent. In addition, we shall prove that those vectors are the span of the vector space $V$, and indeed: $$\text{Span}(\vec{u_1}+\vec{u_2},\vec{u_2}+\vec{u_3},...,\vec{u_{n-2}}+\vec{u_{n-1}},\vec{u_n}+\vec{u_1})=\text{Span}(\vec{u_1},\vec{u_2},...,\vec{u_n})=V$$
$$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare$$

Thoughts: I think that there is an issue with the inverse direction because I haven't used the assumption in which $n$ is odd. I will be grateful if you can tell me what is wrong with the proof, what is good, etc... Thanks!
 A: there are two independant parts in this answer
A direct answer to your questionning
Let us re-formulate your question in the following way:
It is known that a system of $n$ vectors (where $n$ is precisely the dimension of the ambient space), here
$$\left(\overrightarrow{u_{1}} +\overrightarrow{u_{2}}\right),\left(\overrightarrow{u_{2}} +\overrightarrow{u_{3}}\right) \dotsc \left(\overrightarrow{u_{n}} +\overrightarrow{u_{1}}\right)$$
is a basis iff it is an independent system, i.e. a linear combination of its elements is zero iff all these coefficients are zero:
$$\underbrace{\beta _{1}\left(\overrightarrow{u_{1}} +\overrightarrow{u_{2}}\right) +\beta _{2}\left(\overrightarrow{u_{2}} +\overrightarrow{u_{3}}\right) +\dotsc +\beta _{n}\left(\overrightarrow{u_{n}} +\overrightarrow{u_{1}}\right)=0}_{\text{Condition (C)}} \ \implies \ \beta _{1}=\beta _{2}=\cdots \beta _{n}=0$$
Let us write condition (C), as you have done, under the form:
$$( \beta _{1} +\beta _{n})\overrightarrow{u_{1}} +\ ( \beta _{1} +\beta _{2})\overrightarrow{u_{2}} +( \beta _{2} +\beta _{3})\overrightarrow{u_{3}} +\dotsc +( \beta _{n} +\beta _{n-1})\overrightarrow{u_{n}}=0\tag{1}$$
As $\overrightarrow{u_{1}}, \overrightarrow{u_{2}} , \dotsc \overrightarrow{u_{n}}$ is a basis, it is in particular an independant system. Therefore (1) implies:
$$\begin{cases}
\beta _{n} +\beta _{1}&=&0&Eq. 1\\
\beta _{1} +\beta _{2}&=&0&Eq. 2\\
\beta _{2} +\beta _{3}&=&0&Eq. 3\\
&\cdots&&\\
\beta _{n-1} +\beta _{n}&=&0&Eq. n
\end{cases}\tag{2}$$

*

*If $n$ is even, the sum of all equations with odd rank (Eq. 1 + Eq. 3+...) is equal to the sum of all equations with even rank (Eq. 2 + Eq. 4+...). Therefore, the equations (2) aren't independent : we will have an infinity of non zero solutions, all of the form:

$$(\beta_1,\beta_2,... \beta_n)=k(1,-1,1,-1,\cdots 1,-1) \  \text{for any real} \ k$$

*

*if $n$ is odd, let $a:=\beta_n$. Eq. 1 gives $\beta_1=-a$. Eq. 2 gives $\beta_2=a$, etc. till Eq. n giving by an immediate recurrence: $a=(-1)^n a=-a$ which is possible iff $a=0$. As a consequence, we obtain as desired, for any $k$, $\beta_k=0$ !


An alternate answer giving a global view through the use of matrices:
Your change of basis can be written in the following matrix form
$$\begin{pmatrix}|&|&|&\cdots&|&|\\
&&&&&\\
v_1&v_2&v_3&\cdots&v_{n-1}&v_n\\
&&&&&\\
|&|&|&\cdots&|&|\\
\end{pmatrix}=\begin{pmatrix}|&|&|&\cdots&|&|\\
&&&&&\\
u_1&u_2&u_3&\cdots&u_{n-1}&u_n\\
&&&&&\\
|&|&|&\cdots&|&|\\
\end{pmatrix}\underbrace{\begin{pmatrix}1&0&0&\cdots&0&0&1\\
1&1&0&\cdots&0&0&0\\
0&1&1&\cdots&0&0&0\\
\cdots&&&&&\cdots\\
0&0&0&\cdots&1&0&0\\
0&0&0&\cdots&1&1&0\\
1&0&0&\cdots&0&1&1
\end{pmatrix}}_M$$
with a matrix $M$ having a determinant $2 \ne 0$ in the odd case and $0$ in the even case (use a Laplace expansion along the first row).
In the odd case, the inverse matrix entries provides the coefficients to be used for the proof of independence.
