The correct option is A 3Let n= √(6+√(6+√(6+√(6+....)))) n^2= 6+√(6+√(6+√(6+√(6+....)))) n^2=6+n n^2 n 6=0 n^2 3n+2n 6=0 n(n 3)+2( ...

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Integration by Substitution

√6+√6+√6+√6+....

Question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}}$ = ?

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Solution

The correct option is A 3
Let $n = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}}$
$n^{2} = 6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}}$
$n^{2} = 6 + n$
$n^{2} - n - 6 = 0$
$n^{2} - 3 n + 2 n - 6 = 0$
$n (n - 3) + 2 (n - 3)$
$(n + 2) (n - 3)$
$n = (3, - 2)$
Hence $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}} = 3$

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√6+√6+√6+√6+.... = ?

The correct option is A 3Let n=√6+√6+√6+√6+....n2=6+√6+√6+√6+√6+....n2=6+nn2−n−6=0n2−3n+2n−6=0n(n−3)+2(n−3)(n+2)(n−3)n=(3,−2)Hence √6+√6+√6+√6+....=3

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6 + . . . .}}}}$ = ?