SQRT {15\sqrt {15\sqrt {15\sqrt {15\sqrt {15} } } } } \ldots .\infty\)\(\sqrt {n\sqrt {n\sqrt {n\sqrt {n\sqrt n } } } } \ldots .\infty = n\)So,heren = 15∴ x2 + 4 = (15)2 + 4 = 225 + 4 = 229

"> SQRT {15\sqrt {15\sqrt {15\sqrt {15\sqrt {15} } } } } \ldots .\infty\)\(\sqrt {n\sqrt {n\sqrt {n\sqrt {n\sqrt n } } } } \ldots .\infty = n\)So,heren = 15∴ x2 + 4 = (15)2 + 4 = 225 + 4 = 229

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If \(x = \sqrt {15\sqrt {15\sqrt {15\sqrt {15\sqrt {15} } } } } \ldots .\infty \) and x > 0, then find the value of x2 + 4

General Aptitude Algebra in General Aptitude 10 months ago

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Given that\(\SQRT {15\sqrt {15\sqrt {15\sqrt {15\sqrt {15} } } } } \ldots .\infty\)\(\sqrt {n\sqrt {n\sqrt {n\sqrt {n\sqrt n } } } } \ldots .\infty = n\)So,heren = 15∴ x2 + 4 = (15)2 + 4 = 225 + 4 = 229

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