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'Triangles Within Squares' printed from https://nrich.maths.org/
Well explained by Tom, from Cottenham
Village College
The answer I got for $T_n$ is: $8T_n+1=(2n+1)^2$
The $2n+1$ part is because the diagram looks like this for
$T_3$
$2T_n$ form a rectangle $n$ by $n+1$
The four rectangles rotate around the centre and together make
a square of side $n+(n+1)$
So we get the equation $8T_n+1=(2n+1)^2$