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$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ where $a,b$ are natural numbers. Find the value of $a+b$.

I am not able to proceed with solving this question as I have no idea as to how I can calculate $\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}$. A small hint would do.

tc216
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oshhh
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3 Answers3

6

To find $\sqrt{a+\sqrt{a+\cdots}} $, solve the equation

$x = \sqrt{a+x}$

The solution of $\sqrt{31+\sqrt{31+\cdots}}$ gives us $x = \frac{1\pm 5\sqrt{5}}{2}$.

The solution of $\sqrt{1+\sqrt{1+\cdots}}$ gives us $y = \frac{1\pm \sqrt{5}}{2}$.

Thus, we have $$\frac{x}{y} = 6-\sqrt{5}$$ Hence, $a=6, b=5 \Rightarrow a+b = 11$. Hope it helps.

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    @OsheenSachdev See this site : http://math.stackexchange.com/questions/659115/equality-of-nested-radicals-with-different-operations – S.H.W Dec 15 '16 at 11:32
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Hint: To find

$$\sqrt{a+\sqrt{a+\cdots}} $$

solve the equation

$$x = \sqrt{a+x}$$

As for the answer, I get $11$.

Ron Gordon
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√(31+√(31+√(31....))) = s s = √(31+s) s² = s+31 s = (5√5 + 1)/2 (by the quadratic formula)

√(1+√(1+√(1....))) = k k = √(1+k) k² = k+1 k = (√5 + 1)/2 (by the quadratic formula)

s/k = (5√5 + 1)/(√5 + 1) = (5√5 + 1)(√5 - 1)/4 (multiplying the numerator and denominator both by conjugate √5 - 1) = (24-4√5)/4 = 6-√5

Since 6 - √5 = a - √b, a = 6, b = 5

a + b = 11

Thus, the answer is 11.

Mint Tree
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