Problem 40 - Discrete Mathematics with Applications

Question

Using induction, prove each. $$\left(\begin{array}{l}n \\ 0\end{array}\right)^{2}+\left(\begin{array}{l}n \\\ 1\end{array}\right)^{2}+\left(\begin{array}{l}n \\\ 2\end{array}\right)^{2}+\cdots+\left(\begin{array}{l}n \\\ n\end{array}\right)^{2}=\left(\begin{array}{l}2 n \\ n\end{array}\right)$$

Step-by-Step Solution

Answer

By mathematical induction, we successfully proved the identity: $$ \sum_{k=0}^{n}\left(\begin{array}{l}{n} \\\ {k}\end{array}\right)^{2}=\left(\begin{array}{l}{2 n} \\\ {n}\end{array}\right), $$ where the base case (n = 0) and the inductive step are both satisfied.

Step 1: Base Case: n = 0

: For the base case, when n = 0, we can evaluate the sum on the left-hand side: $$ \sum_{k=0}^{0}\left(\begin{array}{l}{0} \\\ {k}\end{array}\right)^{2}=\left(\begin{array}{l}{0} \\\ {0}\end{array}\right)^{2} = 1. $$ On the right-hand side, the binomial coefficient is evaluated as: $$ \left(\begin{array}{l}{2 \cdot 0} \\\ {0}\end{array}\right) = \left(\begin{array}{l}{0} \\\ {0}\end{array}\right) = 1. $$ Since the left-hand side and the right-hand side of the equality are the same, the base case is proved.

Step 2: Induction Hypothesis

: Assume the identity is true for n: $$ \sum_{k=0}^{n}\left(\begin{array}{l}{n} \\\ {k}\end{array}\right)^{2}=\left(\begin{array}{l}{2 n} \\\ {n}\end{array}\right). $$ Now, we must prove it for n + 1.

Step 3: Inductive Step

: Consider the sum for n + 1: $$ \sum_{k=0}^{n+1}\left(\begin{array}{l}{n+1} \\\ {k}\end{array}\right)^{2}. $$ Now we split the sum into two parts: $$ \begin{aligned} &\sum_{k=0}^{n}\left(\begin{array}{l}{n+1} \\\ {k}\end{array}\right)^{2} + \left(\begin{array}{l}{n+1} \\\ {n+1}\end{array}\right)^{2} \end{aligned}. $$ Using the induction hypothesis and the identity \(\left(\begin{array}{l}n+1 \\\\ k\end{array}\right) = \left(\begin{array}{l}n \\\\ k\end{array}\right) + \left(\begin{array}{l}n \\\\ k-1\end{array}\right)\), we can rewrite the sum: $$ \begin{aligned} &\left(\begin{array}{l}{2n} \\\ {n}\end{array}\right) + \sum_{k=0}^{n}\left[\left(\begin{array}{l}n \\\ k\end{array}\right) + \left(\begin{array}{l}n \\\ k-1\end{array}\right)\right]^{2} + 1 \end{aligned}. $$ Now, we expand the squared term and simplify: $$ \begin{aligned} &\left(\begin{array}{l}{2n} \\\ {n}\end{array}\right) + \sum_{k=0}^{n}\left[\left(\begin{array}{l}n \\\ k\end{array}\right)^{2} + 2\left(\begin{array}{l}n \\\ k\end{array}\right)\left(\begin{array}{l}n \\\ k-1\end{array}\right) + \left(\begin{array}{l}n \\\ k-1\end{array}\right)^{2}\right] + 1. \end{aligned}. $$ Now, using \(\left(\begin{array}{l}n \\\\ k\end{array}\right)\left(\begin{array}{l}n \\\\ k-1\end{array}\right) = \frac{n-k+1}{k}\left(\begin{array}{l}n+1 \\\\ k\end{array}\right)\), we substitute in the equalities: $$ \begin{aligned} &\left(\begin{array}{l}{2n} \\\ {n}\end{array}\right) + \sum_{k=0}^{n}\left[\left(\begin{array}{l}n \\\ k\end{array}\right)^{2} + n\left(\begin{array}{l}n+1 \\\ k\end{array}\right) + \left(\begin{array}{l}n \\\ k-1\end{array}\right)^{2}\right] + 1 \end{aligned}. $$ Recognizing the sum terms as the desired formula for n+1, we rewrite the sum: $$ \begin{aligned} &\sum_{k=0}^{n+1}\left(\begin{array}{l}{n+1} \\\ {k}\end{array}\right)^{2}. \end{aligned}. $$ Therefore, the identity is true for n + 1, proving that the inductive step holds. Since both the base case and the inductive step hold, the identity is true for all n by mathematical induction.

Key Concepts

Binomial Coefficients

Binomial coefficients are crucial in various areas of mathematics, particularly in combinatorics. These coefficients are written as \( \binom{n}{k} \) and read as 'n choose k.' They represent the number of ways to choose \( k \) items from \( n \) items without regard for order. This concept is fundamental in calculating combinations and can be evaluated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

where \( n! \) denotes the factorial of \( n \), which is the product of all positive integers up to \( n \). Binomial coefficients appear in the expansion of binomials, particularly as coefficients in Pascal's triangle. These coefficients are symmetrical, meaning \( \binom{n}{k} = \binom{n}{n-k} \). They are incredibly useful in solving problems involving combinations and identities.

Combinatorial Identity

A combinatorial identity is an equation involving binomial coefficients that holds true for all values of parameters involved, often due to the inherent properties of combinations. Such identities reveal unique relationships and symmetries within combinatorial structures. In the exercise, we are asked to prove a given identity involving binomial coefficients:

• \( \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \)

Identities like these are often proved using methods such as generating functions, combinatorial double counting, or mathematical induction, as demonstrated in the provided solution. These techniques show how expressions with seemingly complex sums can reduce to simpler, well-defined forms.

Inductive Hypothesis

The inductive hypothesis is a critical part of mathematical induction, a proof technique used to establish the truth of an infinite sequence of propositions. It involves assuming the proposition is true for some arbitrary case \( n = k \) as a stepping stone to prove it for \( n = k + 1 \). In this context, if the statement is true for one case, it helps prove it true for the successive case, forming a chain of validity. The format usually involves:

• Assuming \( P(k) \) is true (inductive hypothesis)
• Proving that \( P(k+1) \) is true based on the assumption that \( P(k) \) holds

The provided solution uses the inductive hypothesis to transition from proving \( \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n} \) to proving \( \sum_{k=0}^{n+1} \binom{n+1}{k}^2 = \binom{2(n+1)}{n+1} \). This assumption simplifies complex algebraic expressions and utilizes known identities.

Base Case

In mathematical induction, the base case serves as the foundation of the proof. The purpose is to demonstrate that the given statement or formula holds true for an initial value, often \( n = 0 \) or \( n = 1 \). In proving that an identity is true for all integers in a sequence, establishing the base case is crucial.

• For instance, in the exercise, the base case checks if \( \sum_{k=0}^{0} \binom{0}{k}^2 = \binom{0}{0}^2 = 1 \)

This shows the identity holds when \( n = 0 \), effectively starting the chain of induction. If the base case fails, the induction cannot proceed as it breaks the foundation of the entire process. The base case is always verified before proceeding to the inductive hypothesis and inductive step, establishing the base truth from which all subsequent claims are derived.