$ {(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:

or,

Comments?

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

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

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

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?

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

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 ?

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