(20 14? Chenghai district simulation) as shown in the figure, given that ∠ mon = 45, OA 1= 1, make a square a1b1a2 with an area of s1; Make the second square A2B2.

Solution: ∫∠mon = 45,

∴△OA 1B 1 is an isosceles right triangle.

∫OA 1 = 1，

The side length of the square A 1B 1C 1A2 is 1, and the area of the 1 th square is s 1 = 1.

∫b 1c 1∑OA2，

∴∠B2B 1C 1=∠MON=45，

∴△B 1C 1B2 is an isosceles right triangle.

The side length of square A2B2C2A3 is: 1+ 1=2, and the area of the second square is S2=4.

Similarly, the side length of the third square A3B3C3A4 is 2+2=4, and the area of the third square S3 = 16.

The side length of the fourth square A4B4C4A5 is 4+4=8, and the area of the fourth square s 4 is 64.

The side length of the fifth square A5B5C5A6 is 8+8= 16, and the area of the fifth square S5 is 256.

The side length of the sixth square A6B6C6A7 is 16+ 16=32, and the area of the sixth square S6 = 1024.

So the area of the nth square sn = 22n-2.

So the answer is: 64,22n-2.