Question

$\sqrt{6 + \sqrt{6 + \sqrt{6........ \infty } } }$
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Answer 1

$Answer \: \\ \\ Given \: series \: is \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } \\ \\ let \: the \: sum \: of \: this \: be \: = x \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } = x \\ \\ if \: we \: remove \: one \: \sqrt{6} \: the \: series \: will \\ not \: alter. \: becoz \: its \: the \: of \: infinite \: terms. \\ \\ \sqrt{6 + \sqrt{6} + ... + \infty } = x \\ \\ \sqrt{6 + x} = x \\ \\ squaring \: on \: both \: sides \: we \: have \\ \\ ( \sqrt{6 + x} ) {}^{2} = x {}^{2} \\ \\ 6 + x = x {}^{2} \\ \\ x {}^{2} - x - 6 = 0 \\ \\ x {}^{2} - 3x + 2x - 6 = 0 \\ \\ x(x - 3) + 2(x - 3) = 0 \\ \\ (x + 2)(x - 3) = 0 \\ \\ x = - 2 \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: x = 3 \\ \\ x = - 2 \: will \: be \: rejected \: \: becoz \: sum \: of \: this \\ series \: can \:not \: be \: - ive \\ \\ so \: \: \: x = 3 \\ \\ therefore \: sum \: of \: this \: series \: = 3 \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } = 3$

Answer 2

$Answer \: \\ \\ Given \: series \: is \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } \\ \\ let \: the \: sum \: of \: this \: be \: = x \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } = x \\ \\ if \: we \: remove \: one \: \sqrt{6} \: the \: series \: will \\ not \: alter. \: becoz \: its \: the \: of \: infinite \: terms. \\ \\ \sqrt{6 + \sqrt{6} + ... + \infty } = x \\ \\ \sqrt{6 + x} = x \\ \\ squaring \: on \: both \: sides \: we \: have \\ \\ ( \sqrt{6 + x} ) {}^{2} = x {}^{2} \\ \\ 6 + x = x {}^{2} \\ \\ x {}^{2} - x - 6 = 0 \\ \\ x {}^{2} - 3x + 2x - 6 = 0 \\ \\ x(x - 3) + 2(x - 3) = 0 \\ \\ (x + 2)(x - 3) = 0 \\ \\ x = - 2 \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: x = 3 \\ \\ x = - 2 \: will \: be \: rejected \: \: becoz \: sum \: of \: this \\ series \: can \:not \: be \: - ive \\ \\ so \: \: \: x = 3 \\ \\ therefore \: sum \: of \: this \: series \: = 3 \\ \\ \sqrt{6 + \sqrt{6 + \sqrt{6 + ... + \infty } } } = 3$