give the antiderivative, f, of the power series so that f(5) = 0. ∑((-1)^k(x - 5)^k)/(k^k)

Explanation:

Step1: Recall antiderivative rule for power - series

The antiderivative of a power - series $\sum_{k = 1}^{\infty}a_k(x - c)^k$ is $\sum_{k = 1}^{\infty}\frac{a_k}{k + 1}(x - c)^{k+1}+C$. For the power - series $\sum_{k = 1}^{\infty}\frac{(-1)^k(x - 5)^k}{k^k}$, its antiderivative $F(x)$ is $\sum_{k = 1}^{\infty}\frac{(-1)^k}{k^k(k + 1)}(x - 5)^{k + 1}+C$.

Step2: Use the condition $F(5)=0$

Substitute $x = 5$ into $F(x)=\sum_{k = 1}^{\infty}\frac{(-1)^k}{k^k(k + 1)}(x - 5)^{k + 1}+C$. When $x = 5$, $\sum_{k = 1}^{\infty}\frac{(-1)^k}{k^k(k + 1)}(5 - 5)^{k + 1}=0$. So, $F(5)=0 + C$. Since $F(5)=0$, then $C = 0$.

Answer:

$\sum_{k = 1}^{\infty}\frac{(-1)^k}{k^k(k + 1)}(x - 5)^{k + 1}$