Question

Prove that \(\sum_{k=0}^{n}\left(\begin{array}{l}k \\\ r\end{array}\right)=\left(\begin{array}{l}n+1 \\ r+1\end{array}\right),\) where \(1 \leq r \leq n .\)

Short answer

The identity \(\sum_{k=0}^{n}\binom{k}{r} = \binom{n+1}{r+1}\) holds true for all \(n\) where \(1 \leq r \leq n\), as shown by mathematical induction.

Step-by-step solution

Establish base case

The base case for the induction proof is when \(n = r = 1\). For this case, the left side of the equation becomes \(\binom{0}{1} + \binom{1}{1} = 0 + 1 = 1\). The right side of the equation is \(\binom{2}{2} = 1\). Hence, the identity holds for the base case.

Assume statement is true for \(n = p\)

This is the assumption step of the proof. We assume the identity holds for \(n = p\), that is, \(\sum_{k=0}^{p}\binom{k}{r} = \binom{p+1}{r+1}\) holds true.

Prove statement is true for \(n = p + 1\)

We need to show that the identity also holds for \(n = p + 1\) when it holds for \(n = p\). So, we consider the sum \(\sum_{k=0}^{p+1}\binom{k}{r}\). Using our assumption from the previous step, this sum can be rewritten as \(\sum_{k=0}^{p}\binom{k}{r} + \binom{p+1}{r}\). By our assumption this equals \(\binom{p+1}{r+1} + \binom{p+1}{r}\), which further simplifies to \(\binom{p+2}{r+1}\) by the property of binomial coefficients. Thus, the identity holds for \(n = p + 1\) when it holds for \(n = p\).

Conclusion

By principles of mathematical induction, the given identity holds true for all \(n\) where \(1 \leq r \leq n\).

Key concepts

Proof Strategy

When facing mathematical problems, choosing the correct proof strategy is essential. In our case, we've selected mathematical induction, which is perfect for proving statements involving integers, typically those with a recursive nature. Induction consists of two main steps: establishing the base case, and then proving that if the statement holds for a certain case, it also holds for the next.

This method is akin to dominoes falling; the first domino is the base case, and assuming one domino falls (the assumption step for an integer 'p'), ensures that the next domino (integer 'p+1') also falls. This process solidifies the argument step by step, leaving no gaps, which makes mathematical induction a powerful and often used proof strategy in combinatorics and other areas of mathematics.

Binomial Coefficients

Binomial coefficients \( \binom{n}{k} \) are the numbers that arise naturally when expanding a binomial expression \( (a+b)^n \) and they appear as the coefficients. They also have a combinatorial interpretation: \( \binom{n}{k} \) represents the number of distinct ways to choose k items from a collection of n items. This coefficient can be calculated using the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) where \( n! \) denotes the factorial of n.

Furthermore, binomial coefficients obey certain properties, such as symmetry \( \binom{n}{k} = \binom{n}{n-k} \) and Pascal's rule, which states that \( \binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k} \) These properties play a vital role in simplifying complex expressions and are integral to solving combinatorial problems, as seen in our textbook solution.

Combinatorics

Combinatorics pertains to the study of finite or countable discrete structures and involves counting the structures' various features and configurations. It has applications that span across fields such as computer science, operations research, and applied mathematics. Combinatorics covers topics like the counting principle, permutations, combinations, and graph theory.

Why is combinatorics important? It teaches us how to count without counting. For instance, if we want to know how many ways we can arrange a group of people in a line, it's combinatorics that provides the answer without the need to enumerate every possibility. This area of mathematics is essential for understanding probability, and it contributes to the theoretical foundation for statistical inference. Essentially, combinatorics is about exploring and understanding the art of counting and arranging in systematic ways to solve problems.