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.,

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

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,
$$ 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$$
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Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. gauravtiwari.org is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

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7 Comments

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

  3. 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))))
    would you please help me?

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

  5. 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).

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

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