Prove: each square is the sum of 2 consecutive triangular numbers

Is the following an acceptable proof (be harsh (Nod))?

$\displaystyle t_k=1+2+3+\ldots+k=\frac{(k+1)k}{2}$

$\displaystyle t_{k-1}=1+2+3+\ldots+(k-1)=\frac{k(k-1)}{2}$

$\displaystyle \Longrightarrow t_{k-1}+t_k=k^2 \text{ } \square ?$

I mean, is the induction also required? Even if not, I need help with it:

- Base case - $\displaystyle k=1$: $\displaystyle 1^2=t_0+t_1=0+1=1$; define $\displaystyle t_0=0$.

- Induction - $\displaystyle k=k$: suppose $\displaystyle k^2=t_{k-1}+t_k$.

Then $\displaystyle (k+1)^2=t_k+t_{k+1}=?$ ... how do I use the supposition? (Blush)