Prove by induction that sum of an odd number of odd numbers is odd 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.

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? 


*

*we know that the sum is odd for $n=1$.

*knowing the sum is odd for $n=1$ implies that the sum is odd for $n=3$. 

*knowing the sum is odd for $n=3$ implies that the sum is odd for $n=5$.

*knowing the sum if odd for $n=5$ implies that the sum is odd for $n=7$.
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