Don't even think of counting manually

If n is even, the answer would be n(n^2+1)

Here's a stronger result (covers all positive integral values of $n$ ):

Consider the arithmetic progression $(a_k)_{k=1}^\infty$ given by $a_k=2k-1$ . All the successive elements of the triangle follow this AP.

Now, consider the $n^{\textrm{th}}$ row of the triangle. It can be observed that the sum we need is the sum of elements of $(a_k)$ from $k=(T_{n-1}+1)$ to $k=T_n$ where $T_x$ is the $x^{\textrm{th}}$ triangular number . Denote our sum of the elements of the $n^{\textrm{th}}$ row as $\mathcal{S_n}$ . Then,

$\mathcal{S_n}=\sum_{k=T_{n-1}+1}^{T_n} a_k$

Since $(a_k)$ is an AP sequence and we know that $T_x-T_{x-1}=x$ (obviously), we use the summation formulas of AP. We get,

$\large\mathcal{S_n}=\frac{T_n-(T_{n-1}+1)+1}{2}\left(a_{T_{n-1}+1}+a_{T_n}\right)\\ \implies\mathcal{S_n}=\frac n2\bigg(2(T_{n-1}+1)+2(T_n)-2\bigg)\\ \implies \mathcal{S_n}=\frac n2\bigg(2(T_n+T_{n-1})\bigg)=n(2T_{n-1}+n)\\ \implies \mathcal{S_n}=n((n-1)n+n)=n(n(n-1+1))\\ \implies\mathcal{S_n}=n\cdot n\cdot n=n^3$

$\therefore\quad \boxed{\mathcal{S_n}=n^3~\forall~n\in\Bbb{Z^+}}$

Note / Clarification: By definition, we have $T_0=0$ .