find the function represented by the following series and find the interval of convergence of the series. \n…

Explanation:

Step1: Recognize geometric - series form

The given series $\sum_{k = 0}^{\infty}\frac{(x - 5)^{2k}}{4^{k}}=\sum_{k = 0}^{\infty}(\frac{(x - 5)^{2}}{4})^{k}$. A geometric series has the form $\sum_{k=0}^{\infty}ar^{k}$, where $a = 1$ and $r=\frac{(x - 5)^{2}}{4}$.

Step2: Use geometric - series formula

For a geometric series $\sum_{k = 0}^{\infty}ar^{k}$, when $|r|\lt1$, the sum is given by $S=\frac{a}{1 - r}$. Here, since $a = 1$ and $r=\frac{(x - 5)^{2}}{4}$, the function $f(x)=\frac{1}{1-\frac{(x - 5)^{2}}{4}}=\frac{4}{4-(x - 5)^{2}}=\frac{4}{4-(x^{2}-10x + 25)}=\frac{4}{-x^{2}+10x - 21}$.

Step3: Find interval of convergence

For a geometric series $\sum_{k = 0}^{\infty}r^{k}$ to converge, $|r|\lt1$. So, $|\frac{(x - 5)^{2}}{4}|\lt1$. This implies $(x - 5)^{2}\lt4$. Taking the square - root of both sides, we get $|x - 5|\lt2$. Solving the inequality:

• $x-5\lt2$ gives $x\lt7$.
• $-(x - 5)\lt2$ (or $x-5\gt - 2$) gives $x\gt3$.

Answer:

The function $f(x)=\frac{4}{4-(x - 5)^{2}}$ and the interval of convergence is $(3,7)$.