Question

In the expansion of \((x+y)^{n}\), the coefficients of the \(17^{t h}\) and the \(13^{\star}\) terms are equal. Find the number of terms in the expansion. (1) 18 (2) 22 (3) 28 (4) 29

Short answer

#Answer# 26

Step-by-step solution

Binomial Theorem and Coefficients

Recall the binomial theorem, which describes the expansion of \((x+y)^n\) as follows: \((x+y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + ... + \binom{n}{n}y^n \) Here, each term's coefficient is given by the binomial coefficient, \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), where n is the exponent in the expansion and r is a non-negative integer.

Finding the General Form for the 17th and 13th terms

According to the question, the coefficients of the \(17^{th}\) and \(13^{th}\) terms are equal. We know that the \(r^{th}\) term in a binomial expansion has a general form given by: \(T_{r}=\binom{n}{r-1}x^{n-(r-1)}y^{r-1}\) applying this for the 17th and 13th terms: Coefficient of the \(17^{th}\) term = \(\binom{n}{16}\) Coefficient of the \(13^{th}\) term = \(\binom{n}{12}\) Now we will set these coefficients equal to one another and solve for n.

Setting the Coefficients Equal and Solving for n

We have \(\binom{n}{16}=\binom{n}{12}\). By expanding both sides, \(\frac{n!}{16!(n-16)!}=\frac{n!}{12!(n-12)!}\) We can now cancel both sides by n! (which also means n is neither 12 nor 16): \(\frac{1}{16!(n-16)!}=\frac{1}{12!(n-12)!}\) Cross-multiplying, we get: \(12!(n-12)!=16!(n-16)!\) Dividing both sides by \(12!(n-16)!\), we obtain: \(\frac{(n-12!)}{16!}=\frac{13!}{15!}\) Simplifying, we find: \(n-12=13\) Solving for n, we get: \(n=25\)

Finding the Number of Terms

Knowing n, we can now compute the number of terms in the binomial expansion. As there are n+1 terms in the expansion, we simply add 1 to n: Number of terms = n+1 = 25+1 = 26 The correct answer is not in the given options, therefore there might be an error in the given choices.

Key concepts

Binomial Coefficients

When we talk about binomial coefficients, we're referring to the numbers that appear as the coefficients in the binomial theorem expansion. These coefficients play a key role in combinatorics and algebra by providing the number of ways a subset of items can be chosen from a larger set.

To be more specific, a binomial coefficient, noted as \( \binom{n}{r} \), tells us the number of different ways we can pick r unordered outcomes from n possibilities, without repetition. This concept is commonly referred to as 'combinations' in mathematics. The binomial coefficient can also be calculated using the factorial notation as \( \frac{n!}{r!(n-r)!} \), where \( n! \) denotes the factorial of n, which is the product of all positive integers up to n.

Understanding binomial coefficients is essential not only for expanding polynomial expressions but also for solving combinatorial problems like how many different committees or groups can be formed given a set number of people.

Mathematical Induction

Mathematical induction is a powerful technique used to prove the veracity of an infinite sequence of statements. The process of induction involves two main steps. Firstly, we prove that the base case, usually \( P(1) \), is true. Secondly, we make the inductive step, where we assume that a certain statement, \( P(k) \), is true for some arbitrary positive integer k, and then prove that \( P(k+1) \) is also true.

This method can often be seen in action when working with sequences, series, and the properties of numbers. While it wasn't used directly in solving the exercise with the binomial theorem, it's worth noting that many of the properties and formulas we take for granted, including those surrounding the binomial coefficients and their symmetrical nature, can be proved using mathematical induction. For example, proving that \( \binom{n}{r} = \binom{n}{n-r} \), which reflects the symmetry in Pascal's triangle, often involves mathematical induction.

Combinatorics in Algebra

In the realm of algebra, combinatorics enters the picture when we start dealing with combinations and permutations of different elements. The binomial coefficients we explored earlier are a form of combinatorial expression. These coefficients correspond to the combinatorial concept of choosing items—a direct link to combinations.

Combinatorics plays an important role in understanding algebraic structure and expressions. For instance, the expansion of \( (x + y)^n \), as seen in our exercise, is an application of combinatorics. Each term in the expansion represents a combination of x's and y's, multiplied by the appropriate binomial coefficient that signifies the number of ways those terms can be combined.

This crossover between combinatorics and algebra is fascinating, as it allows us to count and manage the possibilities within algebraic manipulations. It can also lead to solving more advanced problems involving probabilities, game theory, and even in planning and decision-making scenarios where numerous possibilities need to be accounted for systematically.