Let's see if we can find an easy pattern. Start with the sequence √(0×1×2×3+1)=1, then continue by increasing the starting point by 1:
√(1×2×3×4+1)=5, √(2×3×4×5+1)=11, √(3×4×5×6+1)=19.
If we call the starting point x and the square root result y we can pair the results (x,y)=(0,1), (1,5), (2,11), (3,19), ...
Now note the difference between consecutive y values: 4, 6, 8, ...
And the second difference is 2 which suggests a quadratic function represented by y=ax2+bx+c.
We need three equations to find a, b, c, so we use the first three pairs. The first pair gives us c=1.
We have two equations left and two unknowns a and b, so we can subtract c from the y values to get (1,4), (2,10).
x=1: 4=a+b;
x=2: 10=4a+2b, which reduces to 5=2a+b. Subtract the other equation: 1=a, so b=4-1=3, and y=x2+3x+1.
To prove this is correct let x=2, then y=4+6+1=11 which matches the point (2,11). When x=3, y=9+9+1=19, matching (3,19).
Therefore, if x=300, y=90000+900+1=90901. This is the solution without using a calculator.