Add, Multiply, Root.

Let us define $a_n = \sqrt{1+n(n+1)(n+2)(n+3)}$ .

$= \sqrt{a^4 + 6a^3 + 11a^2 + 6a + 1}$

$= \sqrt{(a^2 + 3a + 1)^2} \implies a_n = n^2 + 3n + 1$

$a_{204} = (204)^2 + 3(204) + 1 = \boxed{42229}$

If we observe the pattern of the first 4 equations,

Let $a_n$ = $\sqrt{1+n(n+1)(n+2)(n+3)}$

We can see that

n(n+3)+1= $a_n$

So,

$\sqrt{1+204×205×206×207}$

=204×207+1

= $\boxed{42229}$

$a$ = $\sqrt{1+n(n+1)(n+2)(n+3)}$ where $-1 +$ $\displaystyle\sum_{k = 1}^{n+1} 2k$ = $a$ therefore $-1 +$ $\displaystyle\sum_{k = 1}^{205} 2k$ = $42229$ I got this solution by considering these equations as a form of a geometric series with a geometric progression of 2. I'm not sure 100% about this solution, please provide me feedback.

we can clearly see that $a_n = \sqrt{1+n(n+1)(n+2)(n+3)} = (n+1)(n+2)-1$

so, we have $\sqrt{1+204\times205\times206\times207} = 205\times206-1 = \boxed{42229}$