Question

Please quickly solve this:

Answer 1

$HEYA \: \\ \\ let \: the \: sum \: of \: this \: series \: be \: = x \\ \\ x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \sqrt{12} } } } ... \infty \\ \\ if \: we \: remove \: one \: \sqrt{12} \: the \: series \: will \\ not \: alter \: as \: its \: sum \: goes \: up to \: infinity \\ \\ x = \sqrt{12 + x} \\ \\ squaring \: on \: both \: sides \: we \: have \\ \\ x {}^{2} = 12 + x \\ \\ x {}^{2} - x - 12 = 0 \\ \\ x {}^{2} - 4x + 3x - 12 = 0 \\ \\ x(x - 4) + 3(x - 4) = 0 \\ \\ (x + 3) = 0 \: \: or \: \: (x - 4) = 0 \\ \\ x = - 3 \: \: or \: \: x = 4 \\ \\ x = - 3 \: will \: be \: rejected \: beoz \: sum \: \\ of \: numbers \: in \: a \: sqrt \: cant \: be \: negative \: \\ \\ so \: \: x = 4 \: \: is \: only \: possible \: value \\ \\ so \: sum \: of \: this \: series \: is \: 4$