https://r2.erweima.ai/mathbot/image/1726050969864ior78813.jpg

Let's go through the three parts of Exercise 1 in the image you've uploaded:

### 1. Show that for all \( n, p \in \mathbb{N}, n \geq 2, p < n \),
\[ C_n^p = C_{n-1}^{p-1} + C_{n-1}^p \]

This is a basic property of binomial coefficients, known as **Pascal's Rule**. The binomial coefficient is given by:
\[
C_n^p = \binom{n}{p} = \frac{n!}{p!(n-p)!}
\]

Pascal's Rule states:
\[
\binom{n}{p} = \binom{n-1}{p-1} + \binom{n-1}{p}
\]
This rule can be proved combinatorially, where:
- \( \binom{n-1}{p-1} \) counts the number of ways to choose \( p-1 \) elements from \( n-1 \) elements, and we add the \( n \)-th element to form a group of \( p \).
- \( \binom{n-1}{p} \) counts the number of ways to select \( p \) elements from the first \( n

1. Show that for all n, p ∈ N, n ≥ 2, C_n^p = C_{n-1}^{p-1} + C_{n-1}^p. 2. Show that C_n^0 + C_n^1 + ⋯ + C_n^n = 2^n. 3. Calculate the sum: A = Σ_{k=0}^n (2k+1)C_n^k.

Explore combinatorics with a step-by-step solution to a problem involving Pascal's Rule, binomial sums, and the summation formula. This problem demonstrates the use of key combinatorial identities and is suitable for undergraduate math students.

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