Question

If \({ }^{12} \mathrm{C}_{0}{ }^{12} \mathrm{C}_{1}, \ldots . .{ }^{12} \mathrm{C}_{12}\) are the binomial coefficients of the expansion \((1+\mathrm{x})^{12}\), then \({ }^{12} \mathrm{C}_{0}-{ }^{12} \mathrm{C}_{1}+{ }^{12} \mathrm{C}_{2}-{ }^{12} \mathrm{C}_{3} \ldots+{ }^{12} \mathrm{C}_{12}=\) (1) 4096 (2) 1024 (3) 0 (4) \(-1024\)

Short answer

Answer: (3) 0

Step-by-step solution

Recall the binomial theorem and the given expression

The binomial theorem states that for any non-negative integer n and any real numbers a and b: \((a+b)^n = \sum_{k=0}^n { }^n \mathrm{C}_{k} a^{n-k} b^k\) The given expression is: \({ }^{12} \mathrm{C}_{0}-{ }^{12} \mathrm{C}_{1}+{ }^{12} \mathrm{C}_{2}-{ }^{12} \mathrm{C}_{3} \ldots+{ }^{12} \mathrm{C}_{12}\) which can also be written as: \(\sum_{k=0}^{12} (-1)^k { }^{12} \mathrm{C}_{k}\)

Apply the binomial theorem by substituting a=1 and b=-1

Now, we can apply the binomial theorem by substituting a=1 and b=-1: \((1+ (-1))^{12} = \sum_{k=0}^{12} { }^{12} \mathrm{C}_{k} 1^{12-k} (-1)^k\) Simplifying the expression, we obtain: \(0^{12} = \sum_{k=0}^{12} (-1)^k { }^{12} \mathrm{C}_{k}\)

Identify the answer from the given options

Since \(0^{12} = 0\), we can conclude that the value of the alternating sum is: \(\sum_{k=0}^{12} (-1)^k { }^{12} \mathrm{C}_{k} = 0\) From the given options, the correct answer is (3) 0.

Key concepts

Binomial Coefficients

When we talk about binomial coefficients, we often refer to them as the numbers that appear in the expansions of binomial expressions like \((a+b)^n\). They are given by the notation \({ }^n \mathrm{C}_{k}\) and can be found using the combination formula:

• \({ }^n \mathrm{C}_{k} = \frac{n!}{k!(n-k)!}\)

The factorial symbol \(!\) means multiplying a number by each positive integer less than itself. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\).
Binomial coefficients can tell us how many ways we can choose \(k\) elements from \(n\). Each term in the binomial expansion contributes one of these coefficients, showcasing the different combinations of products of \(a\) and \(b\).
In our problem, the expansion of \((1+x)^{12}\) uses these coefficients ranging from \({ }^{12} \mathrm{C}_{0}\) to \({ }^{12} \mathrm{C}_{12}\).

Alternating Series

An alternating series is a series where the signs of the terms alternate between positive and negative. This property plays a crucial role in many mathematical scenarios involving sequence sums.
In the given problem, the expansion involves such an alternating series:

• \(\sum_{k=0}^{12} (-1)^k { }^{12} \mathrm{C}_{k}\)

In this sequence, the \((-1)^k\) factor flips the sign of each term based on whether \(k\) is even or odd. If \(k\) is even, the term remains positive; it becomes negative if \(k\) is odd.
The pattern of flipping signs forms the alternating sum, allowing specific calculations using known properties, like the sum of terms using substitutions. In our case, substituting \(a = 1\) and \(b = -1\) reveals that the entire sum collapses to zero, demonstrating a specific outcome of alternating signs in binomial expansions.

Combinatorics

Combinatorics is a branch of mathematics dealing with counting, arrangement, and combination of objects. It's fundamental when exploring binomial coefficients and their applications.
The problem at hand showcases combinatorics through the calculation of ways to form different groups or sets from given elements without regard to order. This idea is embedded in the structure of binomial coefficients, as these coefficients represent the number of ways to choose \(k\) elements from a set of \(n\). This is central to solving problems in probability and statistics, as well as algebra.
The computation shows how seemingly complicated sequences can resolve into elegant mathematical identities using combination principles, and demonstrates how combinatoric approaches can simplify complex problems, relying on the systematic arrangement and selection of elements to derive meaningful answers.