2. the sum of the series 1 - (π/6)^2 1/2! + (π/6)^4 1/4! - (π/6)^6 1/6! + … + (-1)^n(π/6)^2n/(2n)! is

Explanation:

Step1: Recall cosine series formula

The Mac - laurin series for $\cos x$ is given by $\cos x=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}x^{2n}}{(2n)!}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots+\frac{(-1)^{n}x^{2n}}{(2n)!}$.

Step2: Identify the value of x

In the given series $1 - (\frac{\pi}{6})^{2}\frac{1}{2!}+(\frac{\pi}{6})^{4}\frac{1}{4!}-(\frac{\pi}{6})^{6}\frac{1}{6!}+\cdots+\frac{(-1)^{n}(\frac{\pi}{6})^{2n}}{(2n)!}$, we can see that $x = \frac{\pi}{6}$.

Step3: Calculate the sum

Substitute $x=\frac{\pi}{6}$ into the cosine series formula. So, $\sum_{n = 0}^{\infty}\frac{(-1)^{n}(\frac{\pi}{6})^{2n}}{(2n)!}=\cos(\frac{\pi}{6})$. Since $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$, the sum of the given series is $\frac{\sqrt{3}}{2}$.

Answer:

$\frac{\sqrt{3}}{2}$