we are given the following series and asked to find its sum. \n\n∑(n = 0)^∞ ((-1)^nπ^2n)/(6^2n(2n)!) \n\none…

we are given the following series and asked to find its sum. \n\n∑(n = 0)^∞ ((-1)^nπ^2n)/(6^2n(2n)!) \n\none strategy to find the sum is to rewrite the terms of the series and try to identify similarities to other series we know. \nthe series can be simplified by rewriting the expression with a single power of 2n. \n\n∑(n = 0)^∞ ((-1)^nπ^2n)/(6^2n(2n)!) = ∑(n = 0)^∞ ((-1)^n( )^2n)/(2n)!

we are given the following series and asked to find its sum. \n\n∑(n = 0)^∞ ((-1)^nπ^2n)/(6^2n(2n)!) \n\none strategy to find the sum is to rewrite the terms of the series and try to identify similarities to other series we know. \nthe series can be simplified by rewriting the expression with a single power of 2n. \n\n∑(n = 0)^∞ ((-1)^nπ^2n)/(6^2n(2n)!) = ∑(n = 0)^∞ ((-1)^n( )^2n)/(2n)!

Answer

Explanation:

Step1: Rewrite the series

We know that $\frac{(-1)^n\pi^{2n}}{6^{2n}(2n)!}=\frac{(-1)^n(\frac{\pi}{6})^{2n}}{(2n)!}$ since $\frac{a^m}{b^m}=(\frac{a}{b})^m$. So we can rewrite $\sum_{n = 0}^{\infty}\frac{(-1)^n\pi^{2n}}{6^{2n}(2n)!}$ as $\sum_{n = 0}^{\infty}\frac{(-1)^n(\frac{\pi}{6})^{2n}}{(2n)!}$.

Step2: Recall the cosine - series formula

The Maclaurin series for $\cos x$ is given by $\sum_{n = 0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}$, where $x\in(-\infty,\infty)$.

Step3: Substitute $x$ value

Comparing $\sum_{n = 0}^{\infty}\frac{(-1)^n(\frac{\pi}{6})^{2n}}{(2n)!}$ with the Maclaurin series of $\cos x$, we substitute $x = \frac{\pi}{6}$. Then $\sum_{n = 0}^{\infty}\frac{(-1)^n(\frac{\pi}{6})^{2n}}{(2n)!}=\cos(\frac{\pi}{6})$.

Step4: Evaluate the cosine value

We know that $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$.

Answer:

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