Problem 7
Question
Prove the binomial theorem. $$ \forall n \in \mathbb{N}, \forall x, y \in \mathbb{R},(x+y)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\ k \end{array}\right) x^{n-k} y^{k} $$
Step-by-Step Solution
Verified Answer
By using mathematical induction, it is proven that \((x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k}\) for any natural number \(n\) and any real numbers \(x\) and \(y\).
1Step 1 - Understand the Statement
To prove the binomial theorem, we need to show that for any natural number \(n\) and any real numbers \(x\) and \(y\), the expansion of \((x+y)^n\) equals the sum of \( \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^{k} \)
2Step 2 - Use Mathematical Induction
We'll use induction on \(n\). First, verify the base case where \(n=0\): \( (x+y)^0 = 1 \) and \( \sum_{k=0}^{0} \binom{0}{k} x^{0-k} y^{k} = \binom{0}{0} x^0 y^0 = 1 \). Base case holds true.
3Step 3 - Inductive Step
Assume that the theorem holds for some integer \(n=k\), i.e., \( (x+y)^k = \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} \). Now show it holds for \(n=k+1\).
4Step 4 - Expand \( (x+y)^{k+1} \)
Consider \( (x+y)^{k+1} = (x+y)(x+y)^k \). Using the induction hypothesis, substitute the expansion for \( (x+y)^k \): \( (x+y)^{k+1} = (x+y) \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} \)
5Step 5 - Distribute
Apply distributive property: \( (x+y) \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} = x \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} + y \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} \)
6Step 6 - Simplify
Rewrite the expression: \( x \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} + y \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j} \=\sum_{j=0}^{k} \binom{k}{j} x^{(k+1)-j} y^{j} + \sum_{j=0}^{k} \binom{k}{j} x^{k-j} y^{j+1} \)
7Step 7 - Adjust Indices
Adjust the indices to combine the sums. Let \(i=j\) in the first sum and \(i=j+1\) in the second sum, then combine:\( \sum_{i=0}^{k} \binom{k}{i} x^{(k+1)-i} y^{i} + \sum_{i=1}^{k+1} \binom{k}{i-1} x^{(k+1)-(i-1)} y^{i} \). Now combine:\( x^{k+1} + \sum_{i=1}^{k} \left( \binom{k}{i} + \binom{k}{i-1} \right) x^{k+1-i} y^{i} + y^{k+1} \)
8Step 8 - Binomial Coefficient Property
Use the property of binomial coefficients: \( \binom{k}{i} + \binom{k}{i-1} = \binom{k+1}{i} \). Thus,\( x^{k+1} + \sum_{i=1}^{k} \binom{k+1}{i} x^{k+1-i} y^{i} + y^{k+1} = \sum_{i=0}^{k+1} \binom{k+1}{i} x^{(k+1)-i} y^{i} \)
9Step 9 - Conclusion
Since we have shown \( (x+y)^{k+1} = \sum_{i=0}^{k+1} \binom{k+1}{i} x^{(k+1)-i} y^{i} \), the inductive step is complete, and the binomial theorem holds for all natural numbers \(n\).
Key Concepts
Mathematical InductionBinomial CoefficientsDistributive PropertyNatural Numbers
Mathematical Induction
Mathematical induction is a powerful technique used to prove statements for all natural numbers. It involves two main steps:
By establishing these two steps, we can conclude that the statement is true for all natural numbers.
- Base Case: Show that the statement is true for the initial value (usually n=0 or n=1).
- Inductive Step: Assume the statement is true for an arbitrary natural number k, then prove it's true for k+1.
By establishing these two steps, we can conclude that the statement is true for all natural numbers.
Binomial Coefficients
Binomial coefficients are key in the binomial theorem. They're represented as \( \binom{n}{k} \) and read as 'n choose k'. This represents the number of ways to choose k elements from a set of n elements, given by the formula:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here, ! denotes factorial, which means multiplying a series of descending natural numbers. Binomial coefficients appear in the binomial expansion of \((x + y)^n\), indicating how many ways each term can be formed by choosing powers of x and y.
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Here, ! denotes factorial, which means multiplying a series of descending natural numbers. Binomial coefficients appear in the binomial expansion of \((x + y)^n\), indicating how many ways each term can be formed by choosing powers of x and y.
Distributive Property
The distributive property is fundamental in algebra, and it's used extensively in proofs like the binomial theorem. It states that multiplying a sum by a number gives the same result as multiplying each addend individually and then summing up the products:
\[ a(b + c) = ab + ac \]
This property allows us to distribute terms over addition, simplifying complex expressions into manageable parts.
\[ a(b + c) = ab + ac \]
This property allows us to distribute terms over addition, simplifying complex expressions into manageable parts.
Natural Numbers
Natural numbers, denoted as \( \mathbb{N} \), are the set of positive integers starting from 1 (or 0 in some definitions). These numbers are fundamental in mathematics, particularly in the binomial theorem, which is proved for all natural numbers. They play a critical role in sequences, series, and many algebraic structures.
Other exercises in this chapter
Problem 6
Prove that if 10 points are placed inside an equilateral triangle of side length \(3,\) there will be 2 points within 1 of one another.
View solution Problem 6
State necessary and sufficient conditions for the existence of an Eulerian circuit in a graph.
View solution Problem 7
Prove that in a simple graph (an undirected graph with no loops or parallel edges) having \(n\) nodes, there must be two nodes having the same degree.
View solution Problem 7
State necessary and sufficient conditions for the existence of an Eulerian path in a graph.
View solution