Bearing in mind you asked for different ways:
$(n+1)^2 = n^2 + (2n + 1)$. $2n+1$ is the $n+1$th odd number. (the first odd number is $1 = 2*1 -1$; the second odd number is $3=2*2 -1$ and so on). From that we are inspired to and can easily verify that $m^2 = (m-1)^2 + (2m-1)$. So we can say with confidence that for any positive $m$ that $m^2$ is $(m-1)^2$ plus the $m$ th odd number.
Recursively that means $m^2$ is the sum of the first $m$ odd numbers.
So $\sum_{i=1}^m (2i -1) = m^2$ This the sum of the first $m$ odd numbers.
So $\sum_{i=1}^m (2i) = \sum_{i=1}^m(2i-1) + \sum_{i=1}^m 1= m^2 + m$. This is the sum of the first $m$ even numbers.
So $m^2 + m^2 + m$ is the sum of the first $m$ odd numbers plus the first $m$ even numbers. Or in other words the first $2m$ numbers.
So $\sum_{k = 1}^{2m}k = 2m^2 + m = m(2m + 1)$. Replace $2m$ with $n$ and you get
$\sum_{k=1}^n k= \frac n2(n+1)=\frac {n(n+1)}2$. Which proves it if $n$ is even.
If $n = 2m-1$ is odd then
$\sum_{k = 1}^n k =\sum_{k=1}^{2m-1}k = \sum_{k=1}^{2m}k - 2m = m(2m+1)- 2m = m(2m - 1)$.
Substituting $n = 2m-1$ we have $m = \frac{n+1}2$ and so $m(2m-1) = \frac{n+1}2n=\frac {n(n+1)}2$
That's different.