This problem looks like it involves some clever algebraic manipulation, since computing the product of four consecutive numbers and then adding one is not easy. I notice that the expression under the square root is one more than a product of four numbers, which reminds me of the identity $(a^2-b^2)=(a+b)(a-b).$ Maybe I can rewrite the product of four numbers as a difference of two squares, and then use the identity to simplify the expression. Let me try to find two numbers $a$ and $b$ such that $a^2-b^2=(31)(30)(29)(28).$ I see that the product of the first and last factors is $31 \times 28 = 868,$ and the product of the middle factors is $30 \times 29 = 870.$ These two products differ by two, which means I can write them as $a+b$ and $a-b,$ where $a=\frac{868+870}{2}=869$ and $b=\frac{870-868}{2}=1.$ So I have $a^2-b^2=(869+1)(869-1)=(31)(30)(29)(28).$ Now I can use the identity to write $\sqrt{(31)(30)(29)(28)+1}=\sqrt{a^2-b^2+1}=\sqrt{(a+b)(a-b)+1}.$ Since $a+b=870$ and $a-b=868,$ I can plug them in and get $\sqrt{870 \times 868+1}.$ This looks simpler, but I still need to simplify the expression under the square root. I notice that $870 \times 868+1$ is also one more than a product of two numbers, namely $870 \times 868.$ This means I can use the same identity again, but with different values of $a$ and $b.$ I see that $870 \times 868 = (869^2-1^2),$ so I can write $a=869$ and $b=1.$ Then I have $\sqrt{870 \times 868+1}=\sqrt{(a^2-b^2)+1}=\sqrt{(a+b)(a-b)+1}.$ Plugging in $a+b=870$ and $a-b=868,$ I get $\sqrt{870 \times 868+1}=\sqrt{870 \times 868+1}.$ This is a perfect square, so I can take the square root and get the final answer. 
Compute $\sqrt{(31)(30)(29)(28)+1}.$