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.,

$$x+n+a = \\ \sqrt{ax+{(n+a)}^2+x\sqrt{a(x+n)+{(n+a)}^2+(x+n)\sqrt{a(x+2n)+{(n+a)}^2+(x+2n)\sqrt{\ldots+(x+(k-2)n)\\ \sqrt{a(x+(k-1)n)+{(n+a)}^2+x(x+(k+1)n+a)}}}}}) \\=\sqrt{an+{(x+a)}^2+n\sqrt{a(n+x)+{(x+a)}^2+(n+x)\sqrt{a(n+2x)+{(x+a)}^2+(n+2x)\sqrt{\ldots+(n+(k-2)x) \\ \sqrt{a(n+(k-1)x)+{(x+a)}^2+n(n+(k+1)x+a)}}}}}) \\=\sqrt{xa+{(n+x)}^2+a\sqrt{x(a+n)+{(n+x)}^2+(a+n)\sqrt{x(a+2n)+{(n+x)}^2+(a+2n)\sqrt{\ldots+(a+(k-2)n) \\ \sqrt{ x(a+(k-1)n)+{(n+x)}^2+a(a+(k+1)n+x)}}}}} \ \ldots (10)$$

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:
$$2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{kn+{(n+1)}^2+(k+2)n^2+n}}}}} \ \ldots (18)$$
or,
$$2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$$

You can either start a new conversation or continue an existing one. Please don't use this comment form just to build backlinks. If your comment is not good enough and if in some ways you are trying to just build links — your comment will be deleted. Use this form to build a better and cleaner commenting ecosystem. Students are welcome to ask for help, freebies and more. Your email will not be published or used for any purposes.

1. wow! i’m impressed!…and you’re reading…my blog?

uh…i’m flattered…hope 2012 brings you more math puzzles to solve… :)

2. wow! i’m impressed!…and you’re reading…my blog?

uh…i’m flattered…hope 2012 brings you more math puzzles to solve… :)

3. this is very helpful to me and you are doing a great job . thank you

4. this is very helpful to me and you are doing a great job . thank you

5. hi..i am utkarsh.i have been working on a formula and i am stuck in nested radicals.
basically, i want to find out value of sqrt(2+sqrt(2+sqrt(2…………….sqrt(2)
for x of times,for example, for x=3, i want value of sqrt(2+sqrt(2+sqrt(2+sqrt(2))))

6. hi..i am utkarsh.i have been working on a formula and i am stuck in nested radicals.
basically, i want to find out value of sqrt(2+sqrt(2+sqrt(2…………….sqrt(2)
for x of times,for example, for x=3, i want value of sqrt(2+sqrt(2+sqrt(2+sqrt(2))))

7. by the way,are you left-handed,your hand writing is similiar to mine!

8. by the way,are you left-handed,your hand writing is similiar to mine!

9. Dear Utkarsh! Thanks for reading the post. Before I comment, I would like to mention that Ramanujan Nested Radical formulas are proposed for infinite number of radicals in a number. When, there are finite number of nested radicals, the exact numerical value is calculated by an advanced calculator.
Let me be clear. $sqrt {2}$ always means $sqrt {2}$ or approximately 1.4142… Similarly $sqrt {2+sqrt{2}}$ has its own numerical value. And so on. As we increases the number of squareroots, the value tends to 2 (not exactly 2).
But when infinite terms are considered, the numerical values cam be easily calculated using algebraic equations.
Let $N= sqrt {2+sqrt{2+sqrt{2+ ldots +sqrt{2}}}}$ upto infinte terms
$N= sqrt {2+N}$
or, $N^2-N-2=0$.
The non-negative solution of above quadratic equation is the numerical value of the nested radical (i.e., N=2).

• What is ramanujan redical ?

10. Dear Utkarsh! Thanks for reading the post. Before I comment, I would like to mention that Ramanujan Nested Radical formulas are proposed for infinite number of radicals in a number. When, there are finite number of nested radicals, the exact numerical value is calculated by an advanced calculator.
Let me be clear. $sqrt {2}$ always means $sqrt {2}$ or approximately 1.4142… Similarly $sqrt {2+sqrt{2}}$ has its own numerical value. And so on. As we increases the number of squareroots, the value tends to 2 (not exactly 2).
But when infinite terms are considered, the numerical values cam be easily calculated using algebraic equations.
Let $N= sqrt {2+sqrt{2+sqrt{2+ ldots +sqrt{2}}}}$ upto infinte terms
$N= sqrt {2+N}$
or, $N^2-N-2=0$.
The non-negative solution of above quadratic equation is the numerical value of the nested radical (i.e., N=2).

• What is ramanujan redical ?

11. thanks for the answer!i guess i will really have to use calculators!

12. thanks for the answer!i guess i will really have to use calculators!

$$f(x) = \sqrt{ 1 + x f(x+1) }$$
put $$f(x)=a + bx$$
$$(a + bx)^2 = x^2 + 2x + 1 = (x+1)^2$$
$$a=b=1$$
$$f(x)=x+1$$
$$f(2)=3$$