Well done Tom, from Finham Park School, for clear use of notation :

Whenever you add 3 triangles ( as in

T_{2}

) together with a triangle one size smaller ( as in

T_{1}

), a new triangle is formed (

T_{4}

) , twice the height of the triangle which was used three times (

T_{2}

) .

The smaller triangle can be called $T_{n}$ , while the 3 triangles one size up can be called $T_{n + 1}$ .

One of the $T_{n + 1}$ joins with the $T_{n}$ to form a square of side length $n + 1$ .

The two remaining $T_{n + 1}$ fit to that square producing a large triangle that has a height twice that of $T_{n + 1}$ ,

So the sum of all four triangles is the triangle $T_{2 (n + 1)}$

So $T_{n}$ + $3 T_{n + 1} = T_{2 (n + 1)}$ or, if you prefer, $T_{2 n + 2}$