Well explained by Tom, from Cottenham Village College

The answer I got for

T_{n}

is:

8 T_{n} + 1 = (2 n + 1)^{2}

The $2 n + 1$ part is because the diagram looks like this for $T_{3}$

2 T_{n}

form a rectangle

n

n + 1

The four rectangles rotate around the centre and together make a square of side $n + (n + 1)$

So we get the equation $8 T_{n} + 1 = (2 n + 1)^{2}$