Question

Prove that if \(n=2^{k}-1\) for some \(k \in \mathbb{N}\), then every entry in the \(n\) th row of Pascal's triangle is odd.

Short answer

The statement is proven to be true using the method of mathematical induction and properties of Pascal's Triangle. Thus, it's true that if \(n=2^{k}-1\) for some \(k \in \mathbb{N}\), then every entry in the \(n\) th row of Pascal's triangle is odd.

Step-by-step solution

Base Case

Let us check the base case by setting \(k=1\). This gives us \(n=2^1-1 = 1\). The 1st row of Pascal's Triangle is 1, which is an odd number, hence the base case holds true.

Inductive Hypothesis

Assume that the statement holds true for \(k=p\), where \(p \geq 1\), i.e., every number in the \(2^p-1\)th row is odd.

Inductive Step

We need to prove that the statement holds true for \(k=p+1\), i.e. the \(2^{p+1}-1\) row. Each entry in the Pascal's Triangle can be represented as \(C(n, r)\) where \(n\) is the row number and \(r\) ranges from 0 to \(n\). Using properties of combinations, we know that \(C(n, r-1) + C(n, r) = C(n+1, r)\). Thus \(C(2^{p+1}-1, r) = C(2^p-1, r-1) + C(2^p-1, r)\). The right hand side of the equation are entries from the \(2^p-1\)th row, which are assumed to be odd. The sum of two odd numbers is even. The only for this to hold true is if we carried over during addition, i.e., \(C(2^p-1, r-1) + C(2^p-1, r) = 2m\) for some integer \(m\). But, this will only happen if \(2^p\) divides \(r\), which is not possible as \(r \leq 2^{p+1}-1\).

Conclusion

This contradiction confirms that the \(2^{p+1}-1\) row of the Pascal's Triangle must also consist of odd numbers, thus proving the induction step and hence, proving the whole statement.

Key concepts

Mathematical Induction

Mathematical induction is a powerful proof technique often used to establish the truth of an infinite number of statements. Just like dominoes falling one after another, mathematical induction ensures that if the first case, known as the base case, is true, and if the passage from one case to the next, known as the inductive step, can be confirmed, then the entire sequence of statements must be true.

To apply mathematical induction, we start by verifying the base case, where the statement holds for the smallest value. In this exercise, the base case was checked for when the row number in Pascal's triangle is one, which indeed has all odd entries (simply '1'). Then, we assume that the statement is true for some arbitrary case, called the inductive hypothesis (here, the row number being an exponent of two minus one). Finally, we prove that if the statement is true for this arbitrary case, it must be true for the next case (moving from row number \(2^p-1\) to \(2^{p+1}-1\)). If these steps are successful, we have proven the statement for all natural numbers in the sequence.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It includes studying how to arrange objects according to certain rules and is the core of many problems in probability, statistics, and discrete mathematics.

In the context of Pascal's triangle, combinatorics appears in the form of binomial coefficients. These coefficients count the number of ways to choose a subset of items from a larger set. Pascal's triangle can be visualized as a combinatorial map, where each entry represents a binomial coefficient \(C(n, r)\), corresponding to the number of ways to choose \(r\) elements from a set of \(n\) distinct items. The exercise explores a property of these coefficients when the row number is of a particular form—specifically, one less than a power of two—which relates to the arrangement and counting of elements in combinatorial problems.

Binomial Coefficients

Binomial coefficients are represented as \(C(n, r)\) or \(\binom{n}{r}\) and pronounced as 'n choose r.' These coefficients appear as the numbers in Pascal’s triangle and have an important interpretation in combinatorics: they count the number of ways of selecting \(r\) items from a set of \(n\) distinct items without regard to the order of selection.

Binomial coefficients can be defined recursively using Pascal's triangle, where each entry is the sum of the two entries directly above it. This property can be represented as \(C(n, r-1) + C(n, r) = C(n+1, r)\). The exercise used the characteristics of binomial coefficients to explore the combination of odd numbers and their resulting parity, which was crucial to the problem's solution. When analyzing the odd and even nature of binomial coefficients in specific cases, such as in rows of Pascal's triangle that correspond to one less than powers of two, we dive deep into understanding the patterns and properties of these significant numbers in combinatorics.