Problem 124 - Algebra and Trigonometry Real Mathematics, Real People

Question

Prove the property for all integers \(r\) and \(n,\) where \(0 \leq r \leq n\) \(_{n} C_{0}-_{n} C_{1}+_{n} C_{2}-\cdots \pm_{n} C_{n}=0\)

Step-by-Step Solution

Answer

The given combinatorial property \(_{n} C_{0}-_{n} C_{1}+_{n} C_{2}-\cdots \pm_{n} C_{n}=0\) has been proven to be true by using the Binomial theorem. The theorem was expanded with the given values, and after simplifying the equation, it was found to match with the given property, demonstrating its correctness for all integers \(r\) and \(n\) where \(0 \leq r \leq n\).

Step 1: Binomial Theorem Expansion

To start with the proof, remember the expansion of Binomial theorem: \((a+b)^{n}=\sum_{{r=0}}^{n} {_{n} C_{r} a^{n-r} b^{r}}\) where \(_{n} C_{r}\) represents the number of combinations possible. Now we can apply the values \(a = 1\) and \(b = -1\) to this theorem and expand.

Step 2: Apply the given values

Let's substitute the values of \(a\) and \(b\) into the Binomial theorem expansion: \( (1+(-1))^{n}=\sum_{{r=0}}^{n} {_{n} C_{r} (1)^{n-r} (-1)^{r}}\). This simplifies to \(0^n = \sum_{{r=0}}^{n} {_{n} C_{r}} (-1)^{r}\) as \(1^{n-r}\) gives 1.

Step 3: Simplify the equation

Since anything raised to the power of \(n\) results in 0, excluding when n equals 0, the left side simplifies to 0 except for \(0^{0}\) which is an indeterminate form but is taken as 1 for this context. After expanding the right side, the expression becomes: \(0 =_{n} C_{0}-_{n} C_{1}+_{n} C_{2}- \cdots \pm_{n} C_{n}\).

Step 4: Proof by induction

This exercise does not indicate a required proof by induction, but knowing the property holds true for \(n = 0\) (i.e., \(0 =_{0} C_{0}\)) would suggest it holds for all \(n\). As such, you could, if desired, proceed with a proof by induction.

Key Concepts

Combinations

To understand combinations, think of them as a way to select items from a larger group where the order does not matter. If you have a set of objects and you want to form a subset, a combination helps you count how many such subsets there are without considering any sequence.

In mathematical terms, a combination is represented as \( _{n}C_{r} \), which reads as "n choose r." Here:

• \( n \) is the total number of items.
• \( r \) is the number of items to pick.
• \( _{n}C_{r} \) calculates the number of ways to pick \( r \) items from \( n \) without replacement and without regard to the order.

A simple example would be picking 2 fruits from a set of 4 apples labeled A, B, C, and D. Possible combinations are AB, AC, AD, BC, BD, and CD.

Combinations are fundamental in the binomial theorem and they play a crucial role in calculating coefficients in polynomial expansions.

Binomial Coefficient

The binomial coefficient \( _{n}C_{r} \) is a central element of the binomial theorem. It not only represents combinations but also serves as the coefficient for each term in the binomial expansion.

Mathematically, the binomial coefficient is expressed as:\[ _{n}C_{r} = \frac{n!}{r!(n-r)!} \]where \( n! \) (n factorial) is the product of all positive integers up to \( n \). This formula gives the number of possible combinations when selecting \( r \) items from \( n \).

For example, to find \( _{5}C_{2} \):

• Calculate \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
• Calculate \( 2! = 2 \times 1 = 2 \).
• Calculate \( (5-2)! = 3! = 3 \times 2 \times 1 = 6 \).
• Use the formula: \( _{5}C_{2} = \frac{120}{2 \times 6} = 10 \).

The binomial coefficients also beautifully arrange themselves into Pascal’s Triangle, highlighting the symmetry and structure within binomials.

Algebraic Proof

An algebraic proof involves demonstrating that a mathematical statement or theory is true by using algebraic techniques and logic. In the context of the given exercise, we use the concept of the binomial theorem for the proof.

The proof employs the expression \((a+b)^n\) expanded using the binomial theorem:\[ (a+b)^{n} = \sum_{r=0}^{n} {_{n}C_{r} a^{n-r} b^{r}} \]By substituting \( a = 1 \) and \( b = -1 \), we seek to show the exercise's property:

This substitution simplifies the expression to \((1 + (-1))^n = 0^n = \sum_{r=0}^{n} {_{n}C_{r}} (-1)^r\). Since \(0^n\) equals 0 for all \(n\) (except for the potential \(0^0\)), the right side of the equation becomes:

• \(_{n}C_{0} - _{n}C_{1} + _{n}C_{2} - \cdots \pm _{n}C_{n} = 0\)

This expression confirms the property sought in the proof, showing that the alternating sum of binomial coefficients results in zero.