Prove by induction that if $n$ is odd and $a_1,\,\cdots,\,a_n$ are odd, then $\begin{aligned}\sum_{i = 1}^n a_i\end{aligned}$ is odd.
Progress: If $n = 1$ then $\sum_{i = 1}^1 a_i = a_1$, so the statement is ok. But then I can not understand how to proceed in this case.
On
Proof without induction:
Let $n = 2m+1$ and let $a_i = 2k_i+1$ for $i=1,\cdots,n$. Then
$$\sum_{i=1}^n a_i = \sum_{i=1}^{2m+1} 2k_i+1 = 2\underbrace{\sum_{i=1}^{2m+1} k_i}_{S} + \sum_{i=1}^{2m+1} 1 = 2S + 2m+1 = 2(S+m)+1$$
On
Base case: $a_1$ is odd, therefore it is okay.
Induction hypothesis: For some $k$, $\sum_{i=1}^{2k+1} a_i$ is odd.
Induction step: The sum of two odd numbers is even. Therefore $a_{2k+2}+a_{2k+3}$ is even. And even+odd=odd, and
$$\sum_{i=1}^{2k+3} a_i = \sum_{i=1}^{2k+1} a_i+a_{2k+2}+a_{2k+3}$$
Thus we can conclude it is true.
On
If induction is not required
There is an easy argument. Let your odd number of odd numbers be $a_{1},...,a_{2k+1}$ where $a_{i}=2k_{i}+1$ for each $1\leq i\leq 2k+1$. Then
$$\sum_{i=1}^{2k+1} a_{i} = \sum_{i=1}^{2k+1}(2k_{i}+1)=2\sum_{i=1}^{2k+1}k_{i} + \sum_{i=1}^{2k+1} 1 = 2\left(\sum_{i=1}^{2k+1}k_{i}\right) + 2k+1 = 2\left(k+\sum_{i=1}^{2k+1}k_{i}\right) + 1$$
which is clearly odd since it is one more than a clearly even number.
There is really no need to use alternate indices.
1. Base Case: $n=1$.
$\sum_{i=1}^1 a_i = a_1$ is odd.
2. Induction Hypothesis:
Suppose that $\sum_{i=1}^n a_i $ is odd (for some $n\ge1$, $n$ is odd.)
3. Prove that the statement holds for the next odd number after $n$.
If $n$ is odd, then the next odd number is $n+2$.
$\sum_{i=1}^{n+2} a_i = \underbrace{\sum_{i=1}^{n} a_i}_{\text{first $n$ terms}} + \underbrace{a_{n+1}+a_{n+2}}_{\text{last two terms}}$
By induction hypothesis, $\sum_{i=1}^{n} a_i$ is odd.
$a_{n+1} + a_{n+2}$ is even. (Sum of two odd numbers are even).
So $\sum_{i=1}^{n} a_i + (a_{n+1} + a_{n+2})$ is odd. (Sum of an odd number and an even number is odd.)
Why does this proof work?