## Complete Elementary Analysis of Nested radicals

This is a continuation of the series of summer projects sponsored by department of science and technology, government of India. In this project work, I have worked to collect and expand what Ramanujan did with Nested Radicals and summarized all important facts into the one article. In the article, there are formulas, formulas and only formulas — I think this is exactly what Ramanujan is known for.

This article not only deals with Ramanujan’s initial work on Nested Radicals but also develops few new models and adds more information to it by catching latest research in a very elementary way. The project was initiated first at gauravtiwari.org, about two years ago and done rigorously in recent days.

Download a PDF copy of the article (here) and let me know what you think about it.

## Sequence of real numbers

A sequence of real numbers (or a real sequence) is defined as a function $f: \mathbb{N} \to \mathbb{R}$ , where $\mathbb{N}$ is the set of natural numbers and $\mathbb{R}$ is the set of real numbers. Thus, $f(n)=r_n, \ n \in \mathbb{N}, \ r_n \in \mathbb{R}$ is a function which produces a sequence of real numbers $r_n$ . It’s customary to write a sequence as form of functions in brackets, e.g.; $\langle f(n) \rangle$ , $\{ f(n) \}$ . We can alternatively represent a sequence as the function with natural numbers as subscripts, e.g., $\langle f_n \rangle$ , $\{ f_n \}$ . This alternate method is a better representation of a sequence as it distinguishes ‘a sequence’ from ‘a function’. We shall use $\langle f_n \rangle$ notation and when writen $\langle f_n \rangle$ , we mean $\langle f_1, f_2, f_3, \ldots, f_n, \ldots \rangle$ a sequence with infinitely many terms. Since all of $\{ f_1, f_2, f_3, \ldots, f_n, \ldots \}$ are real numbers, this kind of sequence is called a sequence of real numbers.

## Examples of Sequences

1. Like $f(x)=\dfrac{1}{x} \forall x \in \mathbb{R}$ is a real-valued-function, $f(n)=\dfrac{1}{n} \forall n \in \mathbb{N}$ is a real sequence.

Putting consecutive values of $n \in \mathbb{N}$ in $f(n)=\dfrac{1}{n}$ we obtain a real-sequence

n=1 f(1)=1

n=2 f(2)=1/2

n=3 f(3)=1/3

n=n f(n)=1/n

This real-sequence can be represented by

$\langle \dfrac{1}{n} \rangle := \langle 1, \dfrac{1}{2}, \dfrac{1}{3}, \ldots, \dfrac{1}{n}, \ldots$ .

1. $\langle {(-1)}^n \rangle$ is the sequence $\langle -1, 1, -1, 1, \ldots, {(-1)}^n, \ldots \rangle$ .
2. $\langle -3n \rangle$ is the sequence $\langle -3, -6, -9, \ldots, -3n, \ldots \rangle$
3. A sequence can also be formed by a recurrence relation with boundary values. If $f_n= f_{n-1}+f_{n-2} \ \text{for} n \ge 2$ and $f_0=f_1=1$ , then we obtain the sequence $\langle f_n \rangle$ as
n=1 $f_1=1$ (given)
n=2 $f_2=f_1 +f_0=1+1=2$ (given $f_0=1=f_1$ )
n=3 $f_3=f_2+f_1=2+1=3$
n=4 $f_4=f_3+f_2=3+2=5$
and so on…
This sequence, $\langle 1, 1,2, 3, 5, 8, 13, 21, \ldots \rangle$ is a real-sequence known as Fibonacci Sequence.

## Range Set of a Sequence

The set of all ‘distinct’ elements of a sequence is called the range set of the given sequence.

For example:

• The range set of $\langle \dfrac{1}{n}\rangle:= \{ \dfrac{1}{n} : n \in \mathbb{N} \}$ , which is an infinite set.
• The range set of $\langle {(-1)}^n \rangle := \{ -1, 1 \}$ , a finite set.

Remark: The range set of a sequence may be either infinite or finite, but a sequence has always an infinite number of elements.

## Sub-sequence of the Sequence

A sub-sequence of the sequence is another sequence containing some of the values of the sequence in the same order as in the original sequence. Alternatively, a sub-sequence of the sequence is another sequence which range set is a subset of the range set of the sequence.

For example:

• <1, 3, 5, 7, …> is a sub-sequence of the sequence <1, 2, 3, 4, …>.
• <1, 5, 13, 21, …> is a sub-sequence of the sequence <1,1,2,3,5,8,13,21, 34, …>.
• <1,1,1,1,1,…> is a sub-sequence of the sequence <-1, 1, -1, 1, …>. Since, the sequence <1,1,1,1,…> has only one value for each term, it’s called a constant sequence.

Remark: A sub-sequence is also a sequence hence it satisfy and follow all the properties of a sequence.

## Equality of two sequences

Two sequences $\langle S_n \rangle$ and $\langle T_n \rangle$ are said to be equal, if and only if $S_n=T_n, \forall n \in \mathbb{N}$ .

For example: The sequences $\langle \dfrac{n+1}{n} \rangle$ and $\langle 1+\dfrac{1}{n} \rangle$ are equal to each other.

Remark: From the definition the sequences <-1,1,-1,1, …> and <1,-1,1,-1,…> are not equal to each other, though they look alike and has same range set.

## Algebra of Sequences

Let $\langle S_n \rangle$ and $\langle T_n \rangle$ be two sequence, then the sequences having n-th terms $S_n+T_n, \ S_n-T_n, \ S_n \cdot T_n, \ \dfrac{S_n}{T_n}$ (respectively) are called the SUM, DIFFERENCE, PRODUCT, QUOTIENT of $\langle S_n \rangle$ and $\langle T_n \rangle$ .

For example: The sequence <1, 8, 19,30, …> is the sum of sequences <0, 1, 2, 3, …> and <1, 7, 17, 27, …> obtained after adding n-th term of one sequence to corresponding n-th term of other sequence. Similarly, other operations can be carried.

If $S_n \ne 0 \forall n$ , then the sequence $\langle \dfrac{1}{S_n} \rangle$ is known as the reciprocal of the sequence $\langle S_n \rangle$ .

For example: $\langle \dfrac{1}{1}, \dfrac{-1}{2}, \dfrac{1}{3}, \ldots \rangle$ is the reciprocal of the sequence $\langle 1, -2, 3, \ldots \rangle$ .

Remark: The sequences <-1,1,-1,1, …> and <1,-1,1,-1,…> have their reciprocals equal to the original sequence, hence these are called identity-sequences.

If $c \in \mathbb{R}$ then the sequence with n-th term $cS_n$ is called the scalar multiple of sequence $\langle S_n \rangle$ . This sequence is denoted by $\langle cS_n \rangle$ .

## Bounds of a Sequence

• A sequence $\langle S_n \rangle$ is said to be bounded above, if there exists a real number M such that $S_n \le M, \forall n \in \mathbb{N}$ . M is called an upper bound of the sequence $\langle S_n \rangle$ .
• A sequence $\langle S_n \rangle$ is said to be bounded below, if there exists a real number m such that $S_n \ge m, \forall n \in \mathbb{N}$ . m is called a lower bound of the sequence $\langle S_n \rangle$ .
• A sequence $\langle S_n \rangle$ is said to be bounded, if it is both bounded above and bounded below. Thus, if $\langle S_n \rangle$ is a bounded sequence, there exist two real numbers m & M such that $m \le S_n \le M \forall n \in \mathbb{N}$ .

## SETS

In mathematics, Set is a well defined collection of distinct objects. The theory of Set as a mathematical discipline rose up with George Cantor, German mathematician, when he was working on some problems in Trigonometric series and series of real numbers, after he recognized the importance of some distinct collections and intervals.

Cantor defined the set as a ‘plurality conceived as a unity’ (many in one; in other words, mentally putting together a number of things and assigning them into one box).

Mathematically, a Set $S$ is ‘any collection’ of definite, distinguishable objects of our universe, conceived as a whole. The objects (or things) are called the elements or members of the set $S$ . Some sets which are often pronounced in real life are, words like ”bunch”, ”herd”, ”flock” etc. The set is a different entity from any of its members.

For example, a flock of birds (set) is not just only a single bird (member of the set). ‘Flock’ is just a mathematical concept with no material existence but ‘Bird’ or ‘birds’ are real.

## Representing sets

Sets are represented in two main ways:

## Euler’s (Prime to) Prime Generating Equation

The greatest number theorist in mathematical universe, Leonhard Euler had discovered some formulas and relations in number theory, which were based on practices and were correct to limited extent but still stun the mathematicians. The prime generating equation by Euler is a very specific binomial equation on prime numbers and yields more primes than any other relations out there in number theory. Euler told that the equation $f(x)=x^2+x+k$ yields many prime numbers with the values of x being input from x=0 to $x=k-2$ ; k being a prime.

Let’s see how many primes we can get by using different values of $k$ and $x$:

## Irrational Numbers and The Proofs of their Irrationality

“Irrational numbers are those real numbers which are not rational numbers!”

## Def.1: Rational Number

A rational number is a real number which can be expressed in the form of $\frac{a}{b}$ where $a$ and $b$ are both integers relatively prime to each other and $b$ being non-zero.
Following two statements are equivalent to the definition 1.
1. $x=\frac{a}{b}$ is rational if and only if $a$ and $b$ are integers relatively prime to each other and $b$ does not equal to zero.
2.   $x=\frac{a}{b}&space;\in&space;\mathbb{Q}&space;\iff&space;\mathrm{g.c.d.}&space;(a,b)&space;=1,&space;\&space;a&space;\in&space;\mathbb{Z},&space;\&space;b&space;\in&space;\mathbb{Z}&space;\setminus&space;\{0\}$.

## Def. 2: Relatively Prime Numbers

Two integers $a$ and $b$ are said to be relatively prime to each other if the greatest common divisor of $a$ and $b$ is $1$ .
For example: The pairs (2, 9); (4, 7) etc. are such that each element is relatively prime to other.

## Def. 3: Irrational Number

A real number, which does not fit well under the definition of rational numbers is termed as an irrational number.

A silly question: Let, in the definition of a rational numbers, $a=0$ and $b=8$ , then, as we know $\frac{0}{8}=0$ is a rational number, however $8$ can divide both integers $0$ and $8$ , i.e., $\mathrm{g.c.d.} (0,8) =8$ . (Why?) $\Box$

# Primary ways to prove the irrationality of a real number

It is all clear that any real, if not rational, is irrational. So, in order to prove a (real) number irrational, we need to show that it is not a rational number (i.e., not satisfying definition 1). Most popular method to prove irrationality in numbers, is the Proof by Contradiction, in which we first assume the given (irrational) number to be ‘almost’ rational and later we show that our assumption was untrue. There are many more ways to prove the irrational behavior of numbers but all those are more or less derived from the proof by contradiction.
Some methods which I’ll discuss here briefly are:

1. Pythagorean Approach
2. Using Euclidean Algorithm
3. Power series expansion of special numbers
4. Continued Fraction representation of irrational numbers.

## (1) Pythagorean Approach

This proof is due to Pythagoras and thus called Pythagorean Approach to irrationality. In this approach, we assume a number to be first. Later using the fundamental rules of arithmetic, we make sure whether or not our assumption was true. If our assumption was true, the number we took was rational, otherwise irrational.
For example:

Prove that the number $\sqrt{2}$ is irrational.$Proof: Suppose, to the contrary, that$ \sqrt{2}$is a rational number. Then as according to the definition 1, we can write$ \sqrt{2}=\frac{a}{b} \ldots (1)$where$ a$and$ b$are both integers with$ \mathrm{g.c.d.} (a,b) =1$and$ b \ne 0$. Squaring equation (1),$ 2=\frac{a^2}{b^2}$or,$ a^2=2b^2 \ldots (2)$From the equation (2), we can proceed our proof into two ways: Way I:$ \sqrt{2}$is a positive number, therefore we can assume$ a$and$ b$both to be positive. Since,$ a^2=2b^2$then$ a^2=2b \cdot b$or,$ b|a^2$(read as b divides a squared). Since,$ b$is positive integer,$ b \ge 1$. However,$ b=1$is impossible since corresponding$ a=\sqrt{2}$is not an integer. Thus,$ b > 1$and then according to fundamental theorem of arithmetic, there exists at least one prime$ p> 1$which divides$ b$. Mathematically,$ p|b$but as$ b|a^2$. It is clear that$ p|a^2$. This implies that$ p|a$. Since$ p|a$and$ p|b$, therefore$ \mathrm{g.c.d.}(a,b) \ge p$. So for given number,the greatest common divisor of$ a$and$ b$is not$ 1$, but another prime larger than$ 1$. Thus, it fails to satisfy the definition 1. Thus our claim that$ \sqrt{2}$is rational, is untrue. Therefore,$ \sqrt{2}$is an irrational number. Way II: As a deviation, we can proceed our proof from equation (2) by taking the fact into mind that$ \sqrt{2}$is positive. The number (natural number)$ a$can either be odd or even. Let$ a$be odd, i.e.,$ a=2k+1$where$ k\in \{0,1,2,3, \ldots \}$. Therefore$ a^2$would also be odd. Which contradicts (2), since$ 2b^2$is always even and that equals to$ a^2$. Therefore,$ a$must be an even number. Let$ a=2k$. Putting this into (2) we get,$ 4k^2=2b^2$or,$ b^2=2k^2$or,$ b=\sqrt{2} k \mathrm{g.c.d.}(a,b)=\sqrt{2} \ne 1$. Which is contradiction to our claim. Thus$ \sqrt{2}$is an irrational number. In similar ways, one can prove$ \sqrt{3}$,$ \sqrt{5}$,$ \sqrt{7}$etc. to be irrationals. ## (2) Using Euclidean Algorithm This is an interesting variation of Pythagorean proof. Let$ \sqrt{2}=\frac{a}{b}$with$ \mathrm{g.c.d.}(a,b)=1$, then according to Euclidean Algorithm, there must exist integers$ r$and$ s$, satisfying$ ar+bs=1$. or,$ \sqrt{2} \cdot 1 =\sqrt{2}(ar+bs)$or,$ \sqrt{2}=\sqrt{2}ar +\sqrt{2}bs$or,$ \sqrt{2}=(\sqrt{2}a)r+ (\sqrt{2}b)s$or,$ \sqrt{2}=2br +as$. (From$ \sqrt{2}=a/b$we put$ a=\sqrt{2} b$.) This representation of$ \sqrt{2}$leads us to conclude that$ \sqrt{2}$is an integer, which is completely false. Hence our claim that$ \sqrt{2}$can be written in form of$ \frac{a}{b}$is untrue. Thus,$ \sqrt{2}$is irrational. Similarly, we can use other numbers to prove so. ## (3) Power Series Expansion Some irrational numbers, like$ e$, can be proved to be irrational by expanding them and arranging the terms. Over all, it is another form of proof by contradiction but different from the Pythagorean Approach.$ e$can be defined by the following infinite series:$ e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots$. Suppose, to the contrary, that$ e$is rational, and$ e=\frac{a}{b}$(say) where$ a$and$ b$are positive integers. Then for any$ n>b$and also$ n>1$,$ N=n! \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive, since$ \left({e-(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\ldots+\frac{1}{n!})}\right)$is positive. Or,$ N=n! \left({\dfrac{1}{(n+1)!}+\dfrac{1}{(n+2)!}+\dfrac{1}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{n!}{(n+1)!}+\dfrac{n!}{(n+2)!}+\dfrac{n!}{(n+3)!}+\ldots}\right)$or,$ N=\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+1)(n+2)(n+3)}+\ldots}\right) > 0$. It is clear that$ N$is less than$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right)$. And$ \left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)(n+2)}+\dfrac{1}{(n+2)(n+3)}+\ldots}\right) =\left({\dfrac{1}{(n+1)}+\dfrac{1}{(n+1)}-\dfrac{1}{(n+2)}+\dfrac{1}{(n+2)}-\dfrac{1}{(n+3)}+\ldots}\right) =\dfrac{2}{(n+1)}$. Thus,$ N < \dfrac{2}{(n+1)}<1$. So,$ N$being positive integer is less than$ 1$? It is impossible for any integer. Thus our claim is not true and hence$ e$is irrational. ## (4) Continued Fractions Any number, that can be expressed in form of an infinite continued fraction is always irrational. For example: 1)$ e$can be represented in form of infinite continued fractions, thus$ e$is irrational. 2) Similarly$ \pi$is irrational. 3)$ \sqrt{2}$is also irrational.$ \Box$## The Area of a Disk If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius$ R$is$ pi R^2$. The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.$ The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings — $sum$ them till whole disk is obtained.$Mathematically, we can imagine a ring of with radius$ x$and thickness$ dx$, anywhere in the disk having the same center as disk, calculate its area and then sum up (integrate) it from$ x=0$to$ x=R$. Area of a thin ring is since$ pi x dx$. And after integrating we get, area of disk$ A=2 int_0^R pi x dx$or$ A=pi R^2$.$

There is another approach to achieve the area of a disk, A.

Imagine a disk is made up of a number equal sections or arcs. If there are $n$ number of arcs then interior angle of an arc is exactly $frac{2pi}{n}$ , since $2 pi$ is the total angle at the center of disk and we are dividing this angle into $n$ equal parts. If we join two ends of each sections –we can get $n$ identical triangles in which an angle with vertex O is $frac{2 pi}{n}$ . Now, if we can calculate the area of one such section, we can approach to the area of the disk intuitively. This approach is called the method of exhaustion.

Let, we draw two lines joining center O of the disk and points A & B at circumference. It is clear that OB and OA are the radius of the disk. We joined points A and B in order to form a triangle OAB. Now consider that the disk is made up of n-number of such triangles. We see that there is some area remaining outside the line AB and inside the circumference. If we had this triangle thinner, the remaining area must be lesser.

So, if we increase the number of triangles in disk —-we decrease the remaining areas. We can achieve to a point where we can accurately calculate the area of disk when there are infinitely many such triangles or in other words area of one such triangle is very small. So our plan is to find the area of one triangle —sum it up to n — make $n$ tending to infinity to get the area of disk. It is clear that the sum of areas of all identical triangles like OAB must be either less than or equal to area of the disk. We can call triangles like OAB as inscribed triangles.

Now, if we draw a radius-line OT’, perpendicular to AB at point T and intersecting the circumference at point T’, we can easily draw another triangle OA’B’ as shown in figure. AOB and A’OB’ are inscribed and superscribed triangles of disk with same angle at vertex O. So, it is clear that the angle A’OB’ is equal to the angle AOB. Triangle A’OB’ is larger than the circular arc OAB and circular arc OAB is larger than the inscribed triangle AOB. Also, the sum of areas of triangles identical to OA’B’ is either greater than or equal to area of the disk.

## Triangle Inequality

Triangle inequality has its name on a geometrical fact that the length of one side of a triangle can never be greater than the sum of the lengths of other two sides of the triangle. If $a$ , $b$ and $c$ be the three sides of a triangle, then neither $a$ can be greater than $b+c$ , nor$b$ can be greater than $c+a$ and so $c$ can not be greater than $a+b$ .

Consider the triangle in the image, side $a$ shall be equal to the sum of other two sides $b$ and $c$ , only if the triangle behaves like a straight line. Thinking practically, one can say that one side is formed by joining the end points of two other sides.
In modulus form, $|x+y|$ represents the side $a$ if $|x|$ represents side $b$ and $|y|$ represents side $c$ . A modulus is nothing, but the distance of a point on the number line from point zero.

Ramanujan (1887-1920) discovered some formulas on algebraic nested radicals. This article is based on one of those formulas. The main aim of this article is to discuss and derive them intuitively. Nested radicals have many applications in Number Theory$as well as in Numerical Methods$ .
The simple binomial theorem$of degree 2 can be written as:$ {(x+a)}^2=x^2+2xa+a^2 \ \ldots (1)$Replacing$ a$by$ (n+a)$where$ x, n, a \in \mathbb{R}$, we can have$ {(x+(n+a))}^2= x^2+2x(n+a)+{(n+a)}^2$or,$ {(x+n+a)}^2 =x^2+2xn+2ax+{(n+a)}^2$Arranging terms in a way that$ {(x+n+a)}^2 =ax+{(n+a)}^2+x^2+2xn+ax=ax+{(n+a)}^2+x(x+2n+a)$Taking Square-root of both sides or, $ x+n+a=\sqrt{ax+{(n+a)}^2+x(x+2n+a)} \ \ldots (2)$Take a break. And now think about$ (x+2n+a)$in the same way, as:$ x+2n+a =(x+n)+n+a$. Therefore, in equation (2), if we replace$ x$by$ x+n$, we get$ x+2n+a=(x+n)+n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)((x+n)+2n+a)}$or,$ x+2n+a=\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)} \ \ldots (3)$Similarly,$ x+3n+a=\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)} \ \ldots (4)$and also,$ x+4n+a=\sqrt{a(x+3n)+{(n+a)}^2+(x+3n)(x+5n+a)} \ \ldots (5)$Similarly,$ x+kn+a=\sqrt{a(x+(k-1)n)+{(n+a)}^2+(x+(k-1)n)(x+(k+1)n+a)} \ \ldots (6)$where,$ k \in \mathbb{N}$. Putting the value of$ x+2n+a$from equation (3) in equation (2), we get:$ x+n+a=\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)(x+3n+a)}} \ \ldots (7)$Again, putting the value of$ x+3n+a$from equation (4) in equation (7), we get$ x+n+a =\sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)(x+4n+a)}}} \ \ldots (8)$Generalizing$ the result for $k$ -nested radicals:
$x+n+a =\\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}} \ \ldots (9)$
This is the general formula of Ramanujan Nested Radicals up-to $k$ roots.

Some interesting points
As $x,n$ and $a$ all are real numbers, thus they can be interchanged with each other.
i.e.,

etc.

Putting $n=0$ in equation (9)
we have
$x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots+x\sqrt{ax+{a}^2+x(x+a)}}}}} \ \ldots (11)$
or just, $x+a =\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{ax+{a}^2+x\sqrt{\ldots}}}} \ \ldots (12)$

Again putting $x=1 \ a=0$ in (9)

$1+n =\sqrt{{n}^2+\sqrt{n^2+(1+n)\sqrt{{n}^2+(1+2n)\sqrt{\ldots+(1+(k-2)n)\sqrt{{n}^2+1+(k+1)n}}}}} \ldots (13)$

Putting $x=1 \ a=0$ in equation (8)
$1+n =\sqrt{{n}^2+\sqrt{{n}^2+(1+n)\sqrt{{n}^2+(1+2n)(1+4n)}}} \ \ldots (14)$

Again putting $x=a=n$ =n(say) then
$3n=\sqrt{n^2+4{n}^2+n\sqrt{2n^2+4{n}^2+2n\sqrt{3n^2+4{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn^2+4{n}^2+(k+3)n^2}}}}}$
or, $3n=\sqrt{5{n}^2+n\sqrt{6{n}^2+2n\sqrt{7{n}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+4)n^2+(k+3)n^2}}}}} \ \ldots (15)$

Putting $n=1$ in (15)
$3=\sqrt{5+\sqrt{6+2\sqrt{7+3\sqrt{\ldots+(k-1)\sqrt{(2k+7)}}}}} \ \ldots (16)$

Putting $x=n \in \mathbb{N}$ and $a=0$ in (9) we get even numbers
$2n =\sqrt{{n}^2+n\sqrt{{n}^2+2n\sqrt{{n}^2+3n)\sqrt{\ldots+(k-1)n\sqrt{(k-1)n)+{n}^2+(k+2)n^2}}}}} \ \ldots (17)$

Similary putting $x=n \in \mathbb{N}$ and $a=1$ in (9) we get a formula for odd numbers:
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$
or,
$\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$