Chapter 7: Problem 498
In the expansion of \((\mathrm{x}+\mathrm{y})^{13}\) the co-efficient of \(3 \mathrm{rd}\) term and th terms are equal. (a) 12 (b) 11 (c) 8 (d) 13
Short Answer
Expert verified
None of the given options match the obtained value of 'n', which is 5. There might be a mistake in the given options, or the given data in the question might require a revision.
Step by step solution
01
Find the formula for coefficients using the binomial theorem
According to the binomial theorem, the general term in the expansion of \((x+y)^n\) is given by:
\[T_r = \binom{n}{r-1}x^{n-r+1}y^{r-1}\]
where \(T_r\) represents the \(r^{th}\) term in the expansion.
02
Find the 3rd and 4th terms in the expansion
Given that the coefficients of the 3rd and 4th terms are equal, we need to find the coefficients of these terms using the binomial theorem.
For the 3rd term, \(r=3\):
\[T_3 = \binom{n}{2}x^{n-1}y^2\]
For the 4th term, \(r=4\):
\[T_4 = \binom{n}{3}x^{n-2}y^3\]
03
Equate the coefficients of the 3rd and 4th terms
Since the coefficients of the 3rd and 4th terms are equal, we need to equate the coefficients and solve for 'n':
\[\binom{n}{2} = \binom{n}{3}\]
04
Solve for 'n'
Using the combination formula, we can rewrite the equation as:
\[\frac{n!}{2!(n-2)!} = \frac{n!}{3!(n-3)!}\]
Now, we can simplify and solve for 'n':
\[\frac{1}{2(n-2)} = \frac{1}{3(n-3)}\]
Cross-multiplying, we get:
\[3(n-3) = 2(n-2)\]
Expanding and solving for 'n', we get:
\[3n - 9 = 2n - 4\]
\[n = 5\]
Given the options (a) 12, (b) 11, (c) 8, and (d) 13, none of them match with the obtained value of 'n'. There might be a mistake in the given options, or the given data in the question might require a revision.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Coefficient
A binomial coefficient is a crucial part of the binomial theorem. It is denoted as \( \binom{n}{k} \), where \( n \) and \( k \) are non-negative integers. This notation represents the number of different ways to choose \( k \) elements from a set of \( n \) elements, regardless of order. Sometimes referred to as "n choose k," the binomial coefficient is fundamental in combinatorics.
In the context of a binomial expansion, binomial coefficients determine the weighting of each term, representing the number of ways to select particular powers of \( x \) and \( y \) in the expansion.
- The formula for finding a binomial coefficient is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
- Here, \( n! \) (n factorial) means \( n \times (n-1) \times (n-2) \times \ldots \times 1 \).
- Factorials provide a way to compute the number of possible arrangements of a set.
In the context of a binomial expansion, binomial coefficients determine the weighting of each term, representing the number of ways to select particular powers of \( x \) and \( y \) in the expansion.
Combination Formula
The combination formula is directly related to the binomial coefficients. In mathematics, combinations are different from permutations, as they do not account for the order but only for the selection of elements. The combination formula is given by:
It is the basis for many calculations in probability and statistics, especially when considering problems that involve choosing an orderless subset from a larger set.
In the binomial theorem, these calculations decide the coefficients used in the polynomial expansion. For example, when expanding \((x+y)^n\), you use combination formulas to find the coefficient for each term in the expansion.
- \( C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
- This formula provides the number of ways to select \( k \) items from \( n \) items.
It is the basis for many calculations in probability and statistics, especially when considering problems that involve choosing an orderless subset from a larger set.
In the binomial theorem, these calculations decide the coefficients used in the polynomial expansion. For example, when expanding \((x+y)^n\), you use combination formulas to find the coefficient for each term in the expansion.
Polynomial Expansion
Polynomial expansion involves expressing a power of a binomial as a sum of terms involving coefficients, variables, and integer exponents.
The goal of polynomial expansion is to provide a more easily manageable form for computation.
This is particularly useful in calculus and higher-level mathematics where expressions are too cumbersome in their original binomial form.
Using the binomial theorem, polynomial expansions can be computed methodically and precisely.
- In algebra, the polynomial expansion turns an expression such as \((x+y)^n\) into a sum of terms like \( ax^m y^k \).
- Here, \( a \) is the coefficient determined by the binomial theorem and \( m+k=n \).
The goal of polynomial expansion is to provide a more easily manageable form for computation.
This is particularly useful in calculus and higher-level mathematics where expressions are too cumbersome in their original binomial form.
Using the binomial theorem, polynomial expansions can be computed methodically and precisely.
Binomial Expansion
The binomial expansion is the process of expanding expressions that are raised to a power, specifically expressions in the form of \((x+y)^n\). This expansion results in a multinomial equation where each term is composed of:
The binomial theorem describing this expansion states that:\[\ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \].
Here, each term \( x^{n-k} y^k \) is weighted by a binomial coefficient \( \binom{n}{k} \), determining its contribution to the final polynomial.
Binomial expansion is a powerful tool because it allows the calculation of large powers without directly performing laborious multiplication. Understanding this concept is key to solving many problems in algebra and calculus, as it provides a systematic method to tackle expressions involving powers and roots.
- Binomial coefficients: Derived from the combination formula.
- Polynomial terms: Formed by multiplying powers of \( x \) and \( y \).
The binomial theorem describing this expansion states that:\[\ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \].
Here, each term \( x^{n-k} y^k \) is weighted by a binomial coefficient \( \binom{n}{k} \), determining its contribution to the final polynomial.
Binomial expansion is a powerful tool because it allows the calculation of large powers without directly performing laborious multiplication. Understanding this concept is key to solving many problems in algebra and calculus, as it provides a systematic method to tackle expressions involving powers and roots.