Calculators are for the lazy

Let $x=30$ . Then,

$\sqrt{1+30\cdot 31\cdot 32\cdot 33}$

$=\sqrt{1+x\cdot (x+1)\cdot (x+2)\cdot (x+3)}$

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

$=\sqrt{(x^2+3x+1)^{2}}$

$=x^{2}+3x+1$

Since $x=30$ ,

$x^{2}+3x+1=900+90+1=\boxed{991}$

Note that all positive integers that can be expressed in the form $1+(n)(n+1)(n+2)(n+3)$ are perfect squares.

Also note that the answer is always the second term multiplied by the fourth term plus one, since $x(x+3)+1=x^2+3x+1$

For people who cannot factorize higher order polynomials:

$30 \times 33 = 990$ and $31 \times 32 = 960 + 32 = 992$

The expression can now be written as:

$\sqrt{1+30 \cdot 31 \cdot 32 \cdot 33}\\ =\sqrt{1 + 990 \times 992}$

Let $n=990$ . Then, $992 = n+2$

Substitute them in:

$\sqrt{1 + 990 \times 992}\\ =\sqrt{1+(n)(n+2)}\\ =\sqrt{n^2+2n+1}\\ =\sqrt{(n+1)^2}\\ =n+1 = 990 + 1 = \boxed{991}$

Brute force approach (Please ignore my mistake losing a factor or 10 on 30^4 and 30^3, just showing the method I used) Brute Force

the answer is 30*33+1=991.