Sum of the first $n$ terms of an arithmetic sequence

Write the sum forwards and then backwards, term by term: $S_n = u_n + u_{n-1} + \cdots + u_1$ is exactly the same sum, only listed in reverse order, so writing it directly beneath the forward sum $S_n=u_1+u_2+\cdots+u_n$ and adding the two equations column by column is legitimate.

Look at the $k$-th column of that addition: it is $u_k+u_{n+1-k}$. Since $u_k=u_1+(k-1)d$ and $u_{n+1-k}=u_1+(n-k)d$, adding them gives $u_k+u_{n+1-k}=2u_1+(n-1)d=u_1+u_n$. So every one of the $n$ columns produces the exact same value $u_1+u_n$, regardless of $k$ — this is precisely the cancellation described above.

Summing all $n$ columns therefore gives $2S_n = n(u_1+u_n)$, because the left side is $S_n+S_n$ and the right side is $n$ copies of the constant $u_1+u_n$.

Dividing both sides by $2$ gives $S_n=\dfrac{n(u_1+u_n)}{2}$. Substituting $u_n=u_1+(n-1)d$ into this expression and expanding gives the second form $S_n=nu_1+\dfrac{n(n-1)}{2}d$, which is useful when $u_n$ is not yet known. This argument never divided by anything that could be $0$ and never assumed $n$ is even (the pairing is purely algebraic column-addition, not a physical pairing-up of elements), so it holds for every $n\geq 1$.