For $n=1,2,3,\dots$, let $T_n$ be the $n$-th triangular number, and let $U_n$ be the $n$-th element of our sequence.
We can use the following principle, which we deliberately state a little vaguely. If two sequences begin in the same way, and satisfy the same recurrence, then the sequences are the same. (If the recurrence is second-order, as in the Fibonacci recurrence, "start in the same way" means that the first two terms agree.)
What recurrence shall we use? Note that $U_1=T_1$, so the two sequences start in the same way. It is obvious that $U_{n-1}+U_n=n^2$, or equivalently $U_n=n^2-T_{n-1}$. So if we can show that $T_n=n^2-T_{n-1}$, we will be finished.
In this case, we know an explicit formula for the $n$-th triangular number, since, essentially from the definition of $T_n$, we know that $T_n=1+2+\cdots+n$, and the sum of this arithmetic series is well-known, it is $n(n+1)/2$. Thus $T_n+T_{n-1}=\frac{(n)(n+1)}{2}+\frac{(n-1)(n)}{2}=n^2,$ and therefore the sequence $(T_n)$ satisfies the recurrence $T_n=n^2-T_{n-1}$. This completes the proof.
The general idea is quite useful, since in combinatorics, explicit formulas are (after a while) relatively uncommon, but recurrences are quite common.