If

$$2y = \frac{1}{2^2} + \frac{1 \cdot 3}{2!} \cdot \frac{1}{2^4} + \frac{1 \cdot 3 \cdot 5}{3!} \cdot \frac{1}{2^6} + \dots$$

then prove that $4y^2 + 4y - 1 = 0$.