Quick and cute summer Sangaku #14

One very easy way to solve this problem consists in using coordinate geometry.

• We can draw the first square knowing that a square has right angles.
• We determine the intersection point between $y=x^{2}$ and $y=x$ and we find $P(1;1)$ , we draw the first square and see that its side is egal to $\sqrt{2}$ .
• To draw the second square we determine the intersection point between $y=x^{2}$ and $y=x+2$ and we find $P(2;4)$ , we draw the second square and see that its side is egal to $2\sqrt{2}$ .
• Using a similar method we get the sides of five consecutive squares:
• $s_1=\sqrt{2}$
• $s_2=2\sqrt{2}$
• $s_3=3\sqrt{2}$
• $s_4=4\sqrt{2}$
• $s_5=5\sqrt{2}$
• We can conclude : $s_n=n\sqrt{2}$
• Then $A_n=2n^{2}$
• $\sum _{n=1}^{\infty }\:\frac{1}{2n^2}=\frac{\pi }{12}^2$
• $b-a=12-2=10$