Sum of Infinitely Nested Radicals

Squaring both sides, the equation becomes $x^{2}=k+\sqrt{k+\sqrt{k+\sqrt{k+...}}}$ . You can replace the infinitely nested root with $x$ , making the equation $x^{2}=k+x$ .

By setting the equation equal to 0, it becomes $x^{2}-x-k=0$ . Since $x,k \in \mathbb{N}$ , the equation must be factorable. The center term can only be $-x$ if $k$ is a multiple of two positive consecutive integers $n$ and $n+1$ . The factored form of the equation would be $(x+n)(x-(n+1))=0$ . The solutions would be $x=-n$ or $x=n+1$ , but since $x>1$ , the only solution would be $x=n+1$ .

To find the first five values of $k$ , you must find the first five products of two consecutive positive integers.

• $k=1\cdot2=2$
• $k=2\cdot3=6$
• $k=3\cdot4=12$
• $k=4\cdot5=20$
• $k=5\cdot6=30$

$2+6+12+20+30=\fbox{70}$

We have

$\begin{aligned} & x = \sqrt{k + \sqrt{k + \sqrt{ k+ \cdots}}} \\ \implies & x^2 = k + x \\ \implies & x^2 - x - k = 0 \\ \implies & x = \dfrac{1+\sqrt{1+4k}}{2} & \small\color{#3D99F6}{\text{ As } x > 1} \end{aligned}$

For $x$ to be a natural number $1+4k$ should be the square of an odd integer. Let $1+4k = (2n+1)^2 \implies k = n^2 + n$ . Plugging values of $n$ from $1$ to $5$ , we get $k = 2,6,12,20,30$ , therefore the sum is $\boxed{70}$ .