Infinite triangular Resistance

$It\quad follows\quad from\quad the\quad symmetry\quad considerations\quad \\ that\quad the\quad initial\quad circuit\quad can\quad be\quad replaced\quad by\quad an\quad equivalent\quad one\quad as\quad shown$ $We\quad replace\quad the\quad inner\quad triangle\quad consisting\quad of\quad an\quad infinite\quad number\quad of\quad elements\\ by\quad a\quad resistor\quad of\quad resistance\quad \frac { { R }_{ AB } }{ 2 } \quad ,\quad where\quad the\quad resistance\quad { R }_{ AB }\quad is\quad such\quad that\\ { R }_{ AB }=r=a\rho \quad ,\quad after\quad simplification,\quad the\quad circuit\quad become\quad a\quad system\quad of\quad series\quad and\quad parallel\\ resistances.\quad In\quad order\quad to\quad find\quad r\quad i\quad write\quad the\quad equation\quad \\ r=R\left( R+\frac { R.\frac { r }{ 2 } }{ R+\frac { r }{ 2 } } \right) { \left( R+R+\frac { R.\frac { r }{ 2 } }{ R+\frac { r }{ 2 } } \right) }^{ -1 }\\ solving\quad above\quad equation\quad r=a\rho \left( \frac { \sqrt { 7 } -1 }{ 3 } \right)$