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Magazine Chapter 11: Problem 474
Use the Binomial Theorem to expand \(\left(3 x+\frac{1}{2} y\right)^{6}\)
Short Answer
Expert verified
The expanded form is \(729x^6 + 729x^5y + 303.75x^4y^2 + 37.5x^3y^3 + 1.875x^2y^4 + 0.046875xy^5 + 0.00032552y^6 \).
Step by step solution
01
Write the Binomial Theorem Formula
According to the Binomial Theorem, the expansion of \((a+b)^n\) is given by:\[(a+b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k\] In this exercise, \(a = 3x\), \(b = \frac{1}{2}y\), and \(n = 6\).
02
Identify the First Few Terms
Using the formula from Step 1, the general term in the expansion is given by:\[T_k = \binom{6}{k} (3x)^{6-k} \left(\frac{1}{2}y\right)^k\]We'll compute a few initial terms to begin the expansion.
03
Calculate Coefficient Using the Binomial Coefficient
The binomial coefficient \(\binom{6}{k}\) can be calculated factorially as:\[\binom{6}{k} = \frac{6!}{k!(6-k)!}\]This will be used to calculate the coefficient of each term.
04
Compute the First Term when \(k = 0\)
Using the formula for \(T_k\):\[T_0 = \binom{6}{0} (3x)^6 \left(\frac{1}{2}y\right)^0 = 1 \cdot (3x)^6 = 729x^6\]
05
Compute the Second Term when \(k = 1\)
Again applying the formula to \(k = 1\):\[T_1 = \binom{6}{1} (3x)^5 \left(\frac{1}{2}y\right)^1 = 6 \cdot (243x^5) \cdot \left(\frac{1}{2}y\right)\]This simplifies to:\( 6 \cdot 243x^5 \cdot \frac{1}{2} y = 729 x^5 y\)
06
Compute the Next Terms Incrementally
Repeat the above steps for \(k = 2, 3, \ldots, 6\). For \(k = 2\):\[T_2 = \binom{6}{2} (3x)^4 \left(\frac{1}{2}y\right)^2 = 15 \cdot 81x^4 \cdot \left(\frac{1}{4}y^2\right) = 303.75 x^4 y^2\]Continue this process up to \(k = 6\).
07
Write the Complete Expansion
When you have calculated all terms from \(k = 0\) to \(k = 6\), write out the expansion. It will consist of terms:\( 729x^6 + 729x^5y + 303.75x^4y^2 + 37.5x^3y^3 + 1.875x^2y^4 + 0.046875xy^5 + 0.00032552y^6 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Binomial Coefficient
The binomial coefficient is a vital component used in the Binomial Theorem. It helps calculate the coefficients for each term in a binomial expansion. It is denoted by \(\binom{n}{k}\) and is read as "n choose k." This symbol represents the number of ways to choose \(k\) elements from a set of \(n\) elements. To compute \(\binom{n}{k}\), we use the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]Here's a breakdown:
- \(n!\) (n factorial) is the product of all positive integers up to \(n\).
- \(k!\) is the product of all positive integers up to \(k\).
- (n-k)! is the product of all positive integers up to \((n-k)\).
Expansion of Binomials Using the Binomial Theorem
The expansion of binomials is simplified through the Binomial Theorem. This theorem provides a formula to expand expressions raised to a power. When we say expand \((a+b)^n\), it means expressing this as a sum of terms involving powers of \(a\) and \(b\). The Binomial Theorem states: \[(a+b)^n = \sum_{k=0}^{n}\binom{n}{k} a^{n-k} b^k\]Here's how the process works:
- The sum notation \(\sum_{k=0}^{n}\) indicates you need to calculate terms starting from \(k=0\) up to \(k=n\).
- Each term in the expansion takes the form \(T_k = \binom{n}{k} a^{n-k} b^k\).
- \(a^{n-k}\) diminishes as \(k\) increases while \(b^k\) grows.
Understanding Factorial Calculation
In mathematics, the factorial of a non-negative integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). It is a mathematical concept widely used in permutations, combinations, and algebra. To better understand factorial calculations:
- \(0!\) is defined as 1.
- For any positive integer \(n\), \(n! = n \times (n-1) \times (n-2) \times \ldots \times 1\).
- Factorials grow rapidly with increasing numbers.
