Lemma:
$A \subset \mathbb{R}^n$ is an affine subspace $\Leftrightarrow$ If $\sum \mu_i =1$, and if $a_i \in A$ for any $i =1,...,k$, then we have $\sum \mu _ia_i \in A$.
Definition:
A subspace $A$ in $\mathbb{R}^n$ is an affine subspace if $\lambda a + \mu b \in A$ for any $\lambda, \mu \in \mathbb{R}$ s.t $\lambda + \mu =1$ and for any $a,b \in A$.
I want to prove this lemma using induction (for the direction $\Rightarrow $), but I'm stuck at the last step.
$k=2$ is true by definition of Affine Space.
Assume it is true for $k=n$.
Consider when $k=n+1$.
Let $a_1, ..., a_{n+1} \in A$ be arbitrary elements, and let $\mu_1, ..., \mu_{n+1} \in \mathbb{R}$ are arbitrary numbers that satisfies $\mu_1 +...+\mu_{n+1}=1$.
Observe that if $\mu_n+\mu_{n+1} \neq 0$, $$ \mu_1 a_1+ ...+\mu_{n-1} a_{n-1} + \mu_n a_n + \mu_{n+1} a_{n+1} \\ = \mu_1 a_1+ ...+\mu_{n-1} a_{n-1} + \frac{\mu_n a_n + \mu_{n+1} a_{n+1}}{\mu_n+\mu_{n+1}}(\mu_n + \mu_{n+1})$$
Notice that $$\frac{\mu_n a_n + \mu_{n+1} a_{n+1}}{\mu_n+\mu_{n+1}} \in A$$ by the definition. Thus, by the hypothesis ($k=n$), $\sum \mu_i a_i \in A$.
However, I don't know what to do when $\mu_n + \mu_{n+1}=0$.
Can I get any hint?
Suppose that $n$ is even and that $\mu_i +\mu_j=0$ for all $i$ and $j$. Then $$ 1 = \sum_{i=1}^n\mu_i = \sum_{i=1}^{n/2}(\mu_i +\mu_{i+1})=0\,, $$ which is a contradiction, and so there must be at least one pair of $\mu$'s that don't sum to zero.
Now suppose that $n$ is odd and that $\mu_i +\mu_j=0$ for all $i$ and $j$. Then, $$ 1 = \sum_{i=1}^n\mu_i = \sum_{i=1}^{(n-1)/2}(\mu_i +\mu_{i+1}) + \mu_n =\mu_n\, $$ and so $\mu_n=1$. In this calculation, the choice of pulling out $\mu_n$ was arbitrary, and so the same calculation implies that $\mu_i=1$ for all $i$ (provided that they're all non-zero). This is a contradiction, and hence there must be at least one pair of $\mu$'s that don't sum to zero.
Thus, you can always find two $\mu$'s that don't sum to zero, and without loss of generality, re-order the sum so that they are the last two (i.e., $\mu_n$ and $\mu_{n+1}$), and your proof then works.
(I suspect that there is a more straight-forward way of doing this, directly and without a proof-by-contradiction.)