## Task 1 - 211221

If $a_{1}$ and $d$ are given real numbers, let $\left(a_{n}\right)$ be the arithmetic sequence with $a_{n}=a_{1}+(n-1) d$ for $n=1,2,3, \ldots$. Furthermore, for $n=1,2,3, \ldots$ define:

$$
s_{n}=\sum_{k=1}^{n} a_{k} \quad ; \quad z_{n}=\sum_{k=1}^{n} s_{k}
$$

a) Determine $a_{1}$ and $d$ such that $s_{4}=4$ and $z_{4}=15$.

b) Prove that for arbitrary real $a_{1}, d$ and all $n=1,2,3, \ldots$ the following holds:

$$
z_{n}=\frac{n(n+1)}{2}\left(a_{1}+\frac{n-1}{3} d\right)
$$
b) It is $z_{1}=s_{1}=a_{1}=\frac{1 \cdot 2}{2} \cdot\left(a_{1}+\frac{1-1}{3} \cdot d\right)$, so the statement is true for $n=1$. Further, by induction,

$$
\begin{gathered}
z_{n+1}=z_{n}+s_{n+1}=z_{n}+(n+1) \cdot a_{1}+\frac{n(n+1)}{2} d=\frac{n}{n+1} 2\left(a_{1}+\frac{n-1}{3} d\right)+(n+1) \cdot a_{1}+\frac{n(n+1)}{2} d= \\
=\frac{(n+2)(n+1)}{2} \cdot a_{1}+\frac{n(n+1)}{2} \cdot\left(\frac{n-1}{3}+1\right) d=\frac{(n+1)(n+2)}{2} \cdot\left(a_{1}+\frac{n}{3} d\right)
\end{gathered}
$$

so the statement holds for all natural numbers $n \geq 1$, $\square$.

a) From b) it follows that $15=z_{4}=10 \cdot\left(a_{1}+d\right)$ and $4=s_{4}=4 \cdot a_{1}+6 \cdot d$, thus $a_{1}=\frac{5}{2}$ and $d=-1$. The verification confirms the result.