The tasks presented in this image relate to proving formulas for the partial sums of arithmetic and geometric series using recurrence relations or explicit forms.

Let's go through each of them step-by-step.

Aufgabe 3.6 - Arithmetic Series

The first task asks to show that for an arithmetic sequence, the partial sum $s_n = \sum_{k=1}^n a_k$ is given by:

$s_n = n \cdot \frac{a_1 + a_n}{2}$

where $a_k$ represents terms of an arithmetic sequence, and $n \in \mathbb{N}$.

For an arithmetic series, the $k$-th term $a_k$ can be expressed as: $a_k = a_1 + (k-1)d$ where $d$ is the common difference between terms.

Derivation of the Sum Formula

1. The partial sum of the first $n$ terms can be written as: $s_n = a_1 + a_2 + \dots + a_n$

2. By substituting the terms, we get: $s_n = a_1 + (a_1 + d) + (a_1 + 2d) + \dots + (a_1 + (n-1)d)$

3. Rearranging terms, we can use the formula for the sum of an arithmetic series: $s_n = \frac{n}{2} \cdot (a_1 + a_n)$ This matches the required formula $s_n = n \cdot \frac{a_1 + a_n}{2}$.

4. Additionally, the task specifies the form starting from $k = 0$, which leads to: $s_n = (n + 1) \cdot \frac{a_0 + a_n}{2}$

Aufgabe 3.7 - Geometric Series

The second task involves a geometric series, asking to show that:

$s_n = \sum_{k=1}^n a_k = a_1 \cdot \frac{1 - q^n}{1 - q}$

where $n \in \mathbb{N}$ and $q$ is the common ratio of the series.

For a geometric series, the $k$-th term $a_k$ can be written as: $a_k = a_1 \cdot q^{k-1}$

Derivation of the Sum Formula

1. The sum of the first $n$ terms is: $s_n = a_1 + a_1 q + a_1 q^2 + \dots + a_1 q^{n-1}$

2. This sum of a finite geometric series can be simplified as: $s_n = a_1 \cdot \frac{1 - q^n}{1 - q} \quad \text{for } q \neq 1$

3. Additionally, for the case starting from $k = 0$: $s_n = a_0 \cdot \frac{1 - q^{n+1}}{1 - q}$

Summary

The solutions are:

1. For Aufgabe 3.6 (Arithmetic Series): $s_n = n \cdot \frac{a_1 + a_n}{2} \quad \text{or} \quad s_n = (n + 1) \cdot \frac{a_0 + a_n}{2}$

2. For Aufgabe 3.7 (Geometric Series): $s_n = a_1 \cdot \frac{1 - q^n}{1 - q} \quad \text{or} \quad s_n = a_0 \cdot \frac{1 - q^{n+1}}{1 - q}$

Would you like further details on any specific step?

Related Questions:

1. What is the difference between an arithmetic and a geometric series?
2. How can we derive the formula for the sum of an infinite geometric series?
3. How do these series apply in real-world contexts?
4. Can you explain why the formula for an arithmetic series is linear in $n$ while that for a geometric series is exponential?
5. How would these formulas change if the series started from a different index?

Tip: In arithmetic series, the sum formula relies on the mean of the first and last terms, while in geometric series, it relies on the ratio between successive terms.