Triangular square numbers

Any triangular number can be represented of the form $\frac{n(n+1)}{2}$ (since they are of the form $1+2+3+\ldots+n$ ). Then, the succeeding triangular number will be equal to $\frac{(n+1)(n+2)}{2}$ .

Their sum will be $\frac{n(n+1)+(n+1)(n+2)}{2} \Rightarrow \frac{2(n+1)(n+1)}{2} \Rightarrow (n+1)^2$

Therefore, the sum of any two consecutive triangular numbers is a perfect square.

If you draw $n^2$ dots(an $n\times n$ square),and rotate it 45 degrees.Count it again (row by row,from left to right,up to down),there will be $1+2+3+...+(n-1)+n+(n-1)+...+3+2+1$ dots.We get $n^2=[1+2+...+(n-1)+n]+[1+2+...+(n-2)+(n-1)]$ .Hence,a square is the sum of two consecutive triangular numbers.

Let $T_{k-1}$ and $T_k$ be two consecutive triangular numbers. Now under addition operations we have $\small{ T_{k-1} + T_ k = \dfrac{\,(k-1)\,(k-1 +1)}{2} + \dfrac{k\,(k+1)}{2} = \dfrac{k}{2}\left(k-1 + k+1\right) = k^2 }$ Therefore, the sum of two consecutive triangular number is always a square number.

The much effective solution would be pictorial representation of sum of such two numbers