Problem 29
Question
Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (2 a+b)^{6} $$
Step-by-Step Solution
Verified Answer
\((2a+b)^6 = 64a^6 + 192a^5*b + 240a^4*b^2 + 160a^3*b^3 + 60a^2*b^4 + 12ab^5 + b^6\).
1Step 1: Identify the powers of each binomial term
The binomial theorem follows this pattern: \((x+y)^n = \binom{n}{0}x^n*y^0+ \binom{n}{1}x^{n-1}*y^1+...+\binom{n}{n}x^0*y^n\).\nFor the given expression \((2a+b)^6\), identify \(x = 2a\) and \(y = b\), and the power \(n = 6\).
2Step 2: Apply Binomial Theorem
Now substitute \(x, y\) and \(n\) into the binomial expansion theorem. Thus:\[(2a+b)^6 = \binom{6}{0}(2a)^6*b^0 + \binom{6}{1}(2a)^5*b + \binom{6}{2}(2a)^4*b^2 + \binom{6}{3}(2a)^3*b^3 + \binom{6}{4}(2a)^2*b^4 + \binom{6}{5}(2a)*b^5 + \binom{6}{6}*b^6\]
3Step 3: Simplify Each Term
Calculate each term of the expansion:\[64a^6 + 6*32a^5*b + 15*16a^4*b^2 + 20*8a^3*b^3 + 15*4a^2*b^4 + 6*2a*b^5 + b^6\]
4Step 4: Write the Final Answer
Write the simplified and expanded form of the given binomial. Thus, \((2a+b)^6 = 64a^6 + 192a^5*b + 240a^4*b^2 + 160a^3*b^3 + 60a^2*b^4 + 12ab^5 + b^6\).
Key Concepts
Binomial ExpansionCombinationsAlgebraic Expressions
Binomial Expansion
The binomial expansion is a powerful algebraic tool used to expand expressions raised to a power. When you see an expression like \((x+y)^n\), the binomial theorem helps break it down into a polynomial. This method involves repeatedly applying a pattern to distribute powers to each term in the expression. The key is to express any binomial raised to a power \(n\) using the binomial coefficients and powers of its components.
Here's the general idea:
Here's the general idea:
- Use the binomial coefficients \( \binom{n}{k} \) to determine how many times each term will appear in the expanded expression.
- Each term incorporates a combination of the powers of \(x\) and \(y\), making sure their total power adds up to \(n\).
- Adding these terms gives you the expanded form of the binomial.
Combinations
Combinations are a mathematical concept used to determine the number of ways to choose items from a larger set without regard to the order of selection. In the context of the binomial theorem, combinations are used to calculate the binomial coefficients \( \binom{n}{k} \). These coefficients play a critical role in determining the frequency of each term in a binomial expansion.
The combination formula, \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), helps find these coefficients:
The combination formula, \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \), helps find these coefficients:
- \(n!\) ("n factorial") is the product of all positive integers up to \(n\).
- \(k!\) and \((n-k)!\) similarly represent factorials of smaller numbers.
Algebraic Expressions
An algebraic expression involves numbers, variables, and operations (like addition or multiplication) arranged to express a value or equation. These expressions are the building blocks of algebra used to solve various mathematical problems. Simplifying and manipulating these expressions allow us to solve equations and better understand mathematical relationships.
In the process of expanding a binomial like \((2a+b)^6\), each result is another algebraic expression composed of several terms:
In the process of expanding a binomial like \((2a+b)^6\), each result is another algebraic expression composed of several terms:
- Every term, like \(64a^6\) or \(192a^5b\), is a sum of powers of the variables \(a\) and \(b\).
- Operations like multiplication are used to calculate individual term values, incorporating coefficients and powers.
Other exercises in this chapter
Problem 28
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