Problem 30 - College Algebra

Question

Use the Binomial Theorem to expand each binomial and express the result in simplified form. $$ (a+2 b)^{6} $$

Step-by-Step Solution

Answer

\((a+2b)^{6} = a^{6}+ 12a^{5}b + 60a^{4}b^{2} + 160a^{3}b^{3} + 240a^{2}b^{4} + 192ab^{5} + 64b^{6}\)

Step 1: State the Binomial Theorem

The Binomial theorem states that for any numbers \(a\) and \(b\) and any natural number \(n\), \( (a+b)^{n} = a^{n}+ {n \choose 1} a^{n-1}b + {n \choose 2} a^{n-2}b^{2} + {n \choose 3} a^{n-3}b^{3} + .....+ {n \choose n-1} a^{1}b^{n-1} + b^{n} \). Here, \( {n \choose k} \) denotes the number of ways to choose \(k\) elements from a set of \(n\) elements, and is given by \( {n \choose k}= \frac{n!}{k!(n-k)!} \), where \(n!\) is the factorial of \(n\).

Step 2: Apply the Binomial Theorem

Applying the Binomial theorem for the expression \( (a+2b)^{6} \). The expansion will be: \((a+2b)^{6} = a^{6}+ {6 \choose 1} a^{5}(2b) + {6 \choose 2} a^{4}(2b)^{2} + {6 \choose 3} a^{3}(2b)^{3} + {6 \choose 4} a^{2}(2b)^{4} + {6 \choose 5} a^{1}(2b)^{5} + (2b)^{6}\).

Step 3: Calculation of binomial coefficients and simplification

Now, use the formula for binomial coefficients and simplify terms: \((a+2b)^{6} = a^{6}+ 6a^{5}2^{1}b + 15a^{4}2^{2}b^{2} + 20a^{3}2^{3}b^{3} + 15a^{2}2^{4}b^{4} + 6a^{1}2^{5}b^{5} + 2^{6}b^{6} = a^{6}+ 12a^{5}b + 60a^{4}b^{2} + 160a^{3}b^{3} + 240a^{2}b^{4} + 192ab^{5} + 64b^{6}\)

Key Concepts

Understanding Binomial Expansion

The binomial expansion involves expressing a binomial raised to a power as a sum of terms. This process takes a simple binomial expression like \((a+2b)^{6}\) and unfolds it into a detailed series of expressions. Each term in the expansion follows a specific pattern based on the formula given by the Binomial Theorem.
The expansion consists of multiple terms that include powers of both components of the binomial, \(a\) and \(2b\), multiplied together. The exponents decrease and increase in a coordinated manner across the expansion. For example:

• The first term is \(a^6\), representing the highest power of \(a\) with no \(b\).
• In the second term, \(a^5\) is paired with \((2b)^1\).
• This pattern continues, ending with \((2b)^6\) where \(a\) is absent.

Each coefficient of the terms can be calculated using a formula involving binomial coefficients and factorials, ensuring the precise distribution of both components throughout the expansion.

Exploring Binomial Coefficients

In a binomial expansion, binomial coefficients are crucial as they dictate the number of each term. They are derived using combinations, represented traditionally as \({n \choose k}\), which calculates the number of ways to choose \(k\) items from \(n\) without regard to order.
For example:

• In \((a+2b)^6\), the coefficient for \(a^5(2b)^1\) is \({6 \choose 1} = 6\).
• This coefficient shows the number of different ways one term can appear in the multiplication process.

The formula for binomial coefficients is \({n \choose k} = \frac{n!}{k!(n-k)!}\). This involves factorial operations and ensures the precise calculation needed for each term in the expansion.
Applying these coefficients allows each term of the binomial expansion to be weighted correctly, ensuring a balanced and accurate expansion.

Demystifying Factorials

Factorials are essential in calculating binomial coefficients. A factorial, noted as \(n!\), is the product of all positive integers up to \(n\).
For example:

• \(4! = 4 \times 3 \times 2 \times 1 = 24\).
• \(0!\) is defined as \(1\), which serves as a mathematical convenience for combinations and other calculations.

When calculating binomial coefficients, factorials help determine the precise number of combinations possible. For instance, the coefficient \({6 \choose 2}\) is computed as:\[\frac{6!}{2! \times (6-2)!} = \frac{720}{2 \times 24} = 15\]
This isn't just an abstract calculation; it provides the exact number of needed terms in the binomial expansion, enabling precise representation and simplification.