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.

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. 
