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# Triangles Within Triangles

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

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