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, $\tiny \dpi{150} 2n+1 =\sqrt{n+{(n+1)}^2+n\sqrt{2n+{(n+1)}^2+2n\sqrt{3n+{(n+1)}^2+3n\sqrt{\ldots+(k-1)n\sqrt{(k+3)n^2+(k+3)n+1}}}}} \ \ldots (19)$ Comments? Gaurav Tiwari A designer by profession, a mathematician by education but a Blogger by hobby. With an experience of over seven years with WordPress, PHP and CSS3, Gaurav is capable of doing almost anything related to these. Beyond that, He is a mathematics graduate & a civil service aspirant. #### You may also like... ### 7 Responses 1. hugmamma says: wow! i’m impressed!…and you’re reading…my blog? uh…i’m flattered…hope 2012 brings you more math puzzles to solve… ðŸ™‚ 2. rexantony says: this is very helpful to me and you are doing a great job . thank you 3. utkarsh says: 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. utkarsh says: 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).

• Akshay kumar says:

What is ramanujan redical ?

6. utkarsh says:

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