What's the relationship between the two numbers?

According to the question, $\large \sqrt{18 + 3 \sqrt{ 18 + 3 \sqrt {18 + 3 \sqrt{ 18 + \ldots}}}} = \ x$

Or, $\large \sqrt{18 + 3x} = \ x$

Squaring both the sides , $\large 18+3x= x^{2}$

Or, $\large x^{2}-3x-18=0$

Or, $\large x^{2}-6x+3x-18=0$

Or, $\large x(x-6)+3(x-6)$

Or, $\large (x-6)(x+3)$

Therefore, we get $\large x=6$ or, $\large x=-3$

As the answer can't be negative we can ignore the second value of $x$ .

Therefore, $x=6$ , i.e. $\large \sqrt{18 + 3 \sqrt{ 18 + 3 \sqrt {18 + 3 \sqrt{ 18 + \ldots}}}} = \ 6$

This is just another variant of the classic $\sqrt{x + \sqrt{x + \ldots}}$ problem.

Observe that the expression is the limit of the sequence $\{a_n\}$ where $a_1 = \sqrt{18}, a_{n+1} = \sqrt{18 + 3a_n}$ for all $n \ge 1$ . We claim that:

• $a_n < a_{n+1}$
• $0 < a_n < 6$

by induction on $n$ .

As the base case, $0 < a_1 < 6$ is clear. It can also be computed that $a_2 > a_1$ : since $18 + 3\sqrt{18} > 18$ and the function $x \mapsto \sqrt{x}$ is increasing, $\sqrt{18 + 3\sqrt{18}} > \sqrt{18}$ as well, showing that $a_2 > a_1$ .

Suppose we have proven the claim for $n = n_0 - 1$ , where $n_0 \ge 2$ . We'd like to prove the claim for $n = n_0$ .

We have,

$\begin{aligned} a_{n_0} &= \sqrt{18 + 3 a_{n_0 - 1}} \\ &> \sqrt{18 + 3 \cdot 0} \\ &= \sqrt{18} > 0 \end{aligned}$

Also,

$\begin{aligned} a_{n_0} &= \sqrt{18 + 3 a_{n_0 -1}} \\ &< \sqrt{18 + 3 \cdot 6} \\ &= \sqrt{36} = 6 \end{aligned}$

Finally,

$\begin{aligned} a_{n_0}^2 - a_{n_0-1}^2 &= (18 + 3 a_{n_0 - 1}) - a_{n_0 - 1}^2 \\ &= -(a_{n_0-1} + 3)(a_{n_0-1} - 6) \end{aligned}$

Since $0 < a_{n_0-1} < 6$ by inductive hypothesis, we have $a_{n_0-1}+3 > 0$ and $a_{n_0-1} - 6 < 0$ , giving $-(a_{n_0-1} + 3)(a_{n_0-1} - 6) > 0$ . But $-(a_{n_0-1} + 3)(a_{n_0-1} - 6) = a_{n_0}^2 - a_{n_0-1}^2$ , so $a_{n_0}^2 - a_{n_0-1}^2 > 0$ . This gives $(a_{n_0} + a_{n_0-1})(a_{n_0} - a_{n_0-1}) > 0$ , and since $a_{n_0} + a_{n_0-1} > 0$ , we can divide both sides by $a_{n_0} + a_{n_0-1}$ without changing the direction of the inequality, giving $a_{n_0} - a_{n_0-1} > 0$ or $a_{n_0} > a_{n_0-1}$ . This proves the inductive claim.

Since $\{a_n\}$ is a sequence of real numbers that is increasing and bounded, it has a limit $L$ . This limit satisfies $L = \sqrt{18 + 3L}$ , or,

$\begin{aligned} L^2 &= 18+3L \\ L^2 - 3L - 18 &= 0 \\ (L+3)(L-6) &= 0 \end{aligned}$

This gives $L = -3$ or $L = 6$ . Since $\{a_n\}$ is increasing and $a_1 > 0 > -3$ , clearly the limit cannot be $-3$ , and thus the limit is $\boxed{6}$ .

Let's keep this as simple as we can while still being rigorous.

Consider the continuous function $f(x)=\sqrt{18+3x}$ defined on $[0,6]$ .

We start with a few straightforward observations: (a) $f(x)$ is increasing, (b) $f(6)=6,$ (c) $f(x)>x$ for $x<6$ , (d) The range of $f(x)$ is $[\sqrt{18},6]$ , by properties (a) and (b), so that we can iterate $f(x)$ since the range is a subset of the domain.

Now consider the sequence recursively defined by $a_1=\sqrt{18}$ and $a_{n+1}=f(a_n)$ . By properties (c) and (d), this sequence is increasing and bounded by 6, so that it has a limit $L\leq6$ by completeness. By continuity of $f(x)$ , we have $f(L)$ $=\lim_{n\to\infty}f(a_n)$ $=\lim_{n\to\infty}a_{n+1}=L$ , so that $L=\boxed{6}$ by properties (b) and (c).

We can say that the solution is the last term of the series $a_n = \sqrt{18+3a_{n-1}}; a_0 = 1.$

We proceed by hoping this series converges, and writing a program that iteratively calculates the next value in the series.

import math

a = 1
for i in range(20):
    a = math.sqrt(18 + a * 3)
    print(a)

The values of $a$ converge to 6 at the degree of accuracy of a float.

Let $y=√(18+3√(18+3√(18+3√18+......$ Square RHS and LHS

$y^2=18+3y$

After solving the above quadratic equation, We get $y=6$

LET, the given problem be taken as a function on x where 18+3sqrt(18+.....))))..... be taken as x. now, f(x)= sqrt(18+3sqrt(18+3sqrt)18+....))))))... = sqrt(x) so by property of function any x under sqrt is > or = 0. so x is positive. solving the equation in the form of quadratics we get, x=6 or x=-3. but x is +ve. so x=6

Let x be the required number. Then we note that (x^2)=18+3x, which is the quadratic equation (x^2)-3x-18=0. Factoring the right-hand side of the equation, we obtain (x+3)(x-6)=0 so that x=-3 or x=6. But x cannot take negative values. So x=6 is the only solution. Therefore, the answer is 6.

I'm curious why answer can't be negative. Square root of something came me plus or minus, right?

just guessed

Let x be the expression, therefore

x=√(18+3√(18+3√(18+......

x=√(18+3x)

x² =18+3x

x²-3x-18=0

(x+3)(x-6)=0

x=-3;x=6

Ignoring complex answers(√negative number)

Therefore, x=6