Radical brain warm up

We note that: $\sqrt{1+(n-1)(n+1)} = \sqrt{1 + n^2 - 1} = \sqrt{n^2} = n$ .

Therefore, we have:

$\sqrt{1+2014\sqrt{1+2015\sqrt{1+2016\sqrt{1+2017\times 2019}}}} \\ = \sqrt{1+2014\sqrt{1+2015\sqrt{1+2016\times 2018}}} \\ = \sqrt{1+2014\sqrt{1+2015\times 2017}} \\ = \sqrt{1+2014\times 2016} \\ = \boxed{2015}$

$\begin{aligned} (x+n)^{2} = x^{2} + 2xn + n^{2} \\= x^{2} + n(x + (x+n)) \end{aligned}$

We can continue with this method , but it'll have a limited scope , so to cover a whole new level of possible expressions , let's introduce a new variable k into the expression .

$x+n = \sqrt{ x^{2} + n(x - k + (x + (n+k)))} \\= \sqrt{x^{2} + n(x-k + \sqrt{x^{2} + (n+k)\cdot (x-k + \sqrt{x^{2} + (n+2k)\cdot (x- k + \dots }}}$

Now , we just have to set the values of n=2014 and x=1 .

Hence , the L.H.S becomes 1+2014 = 2015 .