14. (6 points) As shown in the figure, even numbers starting from 2 are arranged in a clockwise right-angle spiral, with 4, 6, 10, 14, 20, 26, 34, ... appearing at the corners of the spiral. Then 2010 $\qquad$ (fill in “will” or “will not”) appear at the corner of the spiral.
【Solution】Solution: From inside to outside, the number of numbers on each circle is $6, 14, 22, 30, 38 \cdots$, forming an arithmetic sequence with a common difference of 8;

Then, when it reaches the 16th circle, the number of numbers on this circle is: $(16-1) \times 8+6=126$;
The total number of numbers is $(6+126) \times 16 \div 2=1056$,
and 2010 is the: $(2010-2) \div 2+1=1005$th number,
On the 16th circle, the largest number is $1056 \times 2=2112$;
At the corner of the upper left, by backtracking 51 numbers, because this circle has 32 numbers in width and 33 numbers in length, $32+33$ $=65$,

Then 2010 is exactly the 20th number from left to right on the bottom side, not reaching the 33rd, in the middle position.
Therefore, it will not appear at the corner.
The answer is: No.