## Task 3 - 271223

a) For each natural number $n$, a function $f$ (with the domain of all real $x \neq 0$) is defined by

$$
f(x)=\sum_{k=0}^{n}(k-2) \cdot x^{k}
$$

Determine all natural numbers $n$ for which the function $f$ satisfies the equation $f(-1)=-f(1)$.

b) For each natural number $n$, a function $g$ (with the same domain) is defined by

$$
g(x)=\sum_{k=0}^{n} \frac{1}{3 k-2} \cdot x^{k}
$$

Investigate whether there exists a natural number $n$ for which the function $g$ satisfies the equation $g(-1)=-g(1)$.
(a) It holds that $f(-1)=-f(1)$ if and only if $f(1)+f(-1)=0$. For every natural number $n$ we have

\[
\begin{aligned}
f(1) & =(-2)+(-1)+0+1+2+\ldots+(n-2) \\
f(-1) & =(-2)-(-1)+0-1+2-+\ldots+(-1)^{n-2}(n-2)
\end{aligned}
\]

From this it follows: If $n$ is even, say $n=2 m$ with a natural number $m$, then

\[
f(1)+f(-1)=2((-2)+0+2+\ldots+(2 m-2))
\]

If $n$ is odd, say $n=2 m+1$ with a natural number $m$, then (for this $m$) the equation (1) also holds. For $m=0,1,2$, the right-hand side of (1) takes the values $2(-2), 2(-2), 2 \cdot 0$. For larger $m$, only positive summands are added. Therefore, $f(1)+f(-1)=0$ holds exactly for $m=2$; thus it is shown: $f(-1)=-f(1)$ holds exactly for $n=4$ and $n=5$.

(b) It holds that $g(-1)=-g(1)$ if and only if $g(1)+g(-1)=0$. For every natural number $n$ we have

\[
\begin{aligned}
g(1) & =\frac{1}{-2}+\frac{1}{1}+\frac{1}{4}+\frac{1}{7}+\ldots+\frac{1}{3 n-2} \\
g(-1) & =\frac{1}{-2}-\frac{1}{1}+\frac{1}{4}-\frac{1}{7}+-\ldots+(-1)^{n-2} \cdot \frac{1}{3 n-2}
\end{aligned}
\]

From this it follows: If with a natural number $m$ either $n=2 m$ or $n=2 m+1$, then

\[
g(1)+g(-1)=2\left(\frac{1}{-2}+\frac{1}{4}+\frac{1}{10}+\ldots+\frac{1}{6 m-2}\right)=-1+\frac{1}{2}+\frac{1}{5}+\ldots+\frac{1}{3 m-1}
\]

For $m \leq 5$ we now have

\[
\frac{1}{2}+\frac{1}{5}+\ldots+\frac{1}{3 m-1} \leq \frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+\frac{1}{14}\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{20}=1
\]

Thus, $g(1)+g(-1)>0$.

Therefore, there is no natural number $m$ with $g(1)+g(-1)=0$ and consequently no natural number $n$ with $g(-1)=-g(1)$.

Adapted from $[5]$