$\frac{\sqrt{31+\sqrt{31+\sqrt{31+ \cdots}}}}{\sqrt{1+\sqrt{1+ \sqrt{1+ \cdots}}}}=a-\sqrt b$ 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.
On
Hint: To find
$$\sqrt{a+\sqrt{a+\cdots}} $$
solve the equation
$$x = \sqrt{a+x}$$
As for the answer, I get $11$.
On
√(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.
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.