Question
$√{1+√{1+√{1 +...}}}$
Answer: Option A
Answer: (a)Let x = $√{1+√{1+√{1 +...}}}$On squaring both sides$x^2 =1+√{1+√{1+√{1 +...}}}$$x^2$ = 1 + x$x^2$ - x - 1 = 0$x = {+1± √{1+4}}/2 = {+1 ±√ 5}/2$But sum of + ve numbers can't be negative.$x={1+ √5}/2 ={1+ 2.236}/2$= ${3.236}/2 =1.618$Thus 1 < 1.618 < 2
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Answer: (a)Let x = $√{1+√{1+√{1 +...}}}$On squaring both sides$x^2 =1+√{1+√{1+√{1 +...}}}$$x^2$ = 1 + x$x^2$ - x - 1 = 0$x = {+1± √{1+4}}/2 = {+1 ±√ 5}/2$But sum of + ve numbers can't be negative.$x={1+ √5}/2 ={1+ 2.236}/2$= ${3.236}/2 =1.618$Thus 1 < 1.618 < 2
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