https://file.aiquickdraw.com/mathbot/file/1726707854981muau36r1.jpg

Let's break down the problem you provided step by step, using **Lucas's Theorem** to help prove the modular binomial coefficients.

### Part (a): Writing the first 9 rows of Pascal's Triangle modulo 2

When reducing Pascal's Triangle modulo 2, every entry is either \(0 \mod 2\) or \(1 \mod 2\). The Pascal's Triangle modulo 2 reveals patterns where the odd numbers become 1 and the even numbers become 0. Here's the first 9 rows:

1. Row 0: \( 1 \)
2. Row 1: \( 1 \quad 1 \)
3. Row 2: \( 1 \quad 0 \quad 1 \)
4. Row 3: \( 1 \quad 1 \quad 1 \quad 1 \)
5. Row 4: \( 1 \quad 0 \quad 0 \quad 0 \quad 1 \)
6. Row 5: \( 1 \quad 1 \quad 0 \quad 0 \quad 1 \quad 1 \)
7. Row 6: \( 1 \quad 0 \quad 1 \quad 0 \quad 1 \quad 0 \quad 1 \)
8. Row 7: \( 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \)
9. Row 8: \( 1 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad 1 \)

This shows how the even and odd terms alternate when reduced modulo 2, forming fractal-like patterns. The resulting rows consist only of binary values (0 and 1).

### Part (b): Proving the two modulo 2 conditions using **Lucas's Theorem**

Lucas's Theorem helps us calculate binomial coefficients modulo a prime number. Specifically, it states:

\[
\binom{n}{k} \mod p = \prod \binom{n_i}{k_i} \mod p
\]

where \( n_i \) and \( k_i \) are the base \(p\)-expansions (in this case base 2 for \( p = 2 \)) of \(n\) and \(k\), respectively.

1. **Proof of the first condition**:
   \[
   \binom{2^n - 1}{k} \equiv 1 \mod 2 \quad \text{for } 0 \leq k \leq 2^n - 1
   \]

   Using Lucas's Theorem, expand \( 2^n - 1 \) and \( k \) in binary. The binary representation of \( 2^n - 1 \) is all ones (\(111\ldots1\)) for \( n \) digits. For any \( k \), the binomial coefficient modulo 2 will be 1 because the binomial coefficient \(\binom{1}{k_i}\) is 1 for \(k_i = 0\) or \(1\) (and 0 otherwise). Therefore:

   \[
   \binom{2^n - 1}{k} \equiv 1 \mod 2
   \]
   for all \( k \) in this range.

2. **Proof of the second condition**:
   \[
   \binom{2^n}{k} \equiv 
   \begin{cases}
   1 \mod 2 & \text{for } k = 0 \text{ or } k = 2^n, \\
   0 \mod 2 & \text{otherwise}.
   \end{cases}
   \]

   Using Lucas's Theorem again, the binary expansion of \( 2^n \) is \( 100\ldots0 \) (a 1 followed by \( n \) zeros). For \( k = 0 \) or \( k = 2^n \), the binomial coefficient is 1 modulo 2. However, for any \( 0 < k < 2^n \), the binomial coefficient will contain zeros in its binary expansion, leading to a product that is zero modulo 2. Hence:

   \[
   \binom{2^n}{k} \equiv 0 \mod 2 \quad \text{for } 0 < k < 2^n.
   \]

This completes the proofs for the two modular binomial coefficients using Lucas's Theorem.

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Let me know if you would like any further clarifications, or if you have any questions on this explanation.

Here are 5 related questions that expand on this material:

1. How does Pascal’s Triangle behave modulo primes larger than 2?
2. Can you derive a pattern in Pascal’s Triangle modulo 3 similar to what we see modulo 2?
3. How is Lucas’s Theorem used to simplify binomial coefficient computations for other moduli?
4. How can you generalize Lucas's Theorem for higher powers of 2, e.g., modulo \(2^m\)?
5. What role does Pascal’s Triangle play in combinatorics, especially for generating functions?

**Tip:** When working with binomial coefficients modulo primes, always check if the theorem applies directly or if adaptations are needed based on the binary (or base-p) expansions of the numbers involved!

Prove that for n = 0, 1, 2, ... (2^n - 1 choose k) ≡ 1 mod 2 for 0 ≤ k ≤ 2^n - 1 and (2^n choose k) ≡ 1 mod 2 for k = 0, 2^n and 0 mod 2 otherwise.

Explore the proof of binomial coefficients modulo 2 using Lucas's Theorem. This involves examining Pascal's triangle modulo 2 and proving properties for (2^n - 1 choose k) and (2^n choose k). Suitable for undergraduate combinatorics and number theory students.

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