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Magazine Chapter 11: Problem 8
Use Pascal's triangle to expand the expression. $$(1+\sqrt{2})^{6}$$
Short Answer
Expert verified
(1 + \sqrt{2})^6 = 99 + 70\sqrt{2}
Step by step solution
01
Understand Pascal's Triangle
Pascal's Triangle is a triangular array of numbers where each entry is the sum of the two directly above it. The rows of Pascal鈥檚 Triangle correspond to the coefficients needed to expand a binomial expression using the Binomial Theorem.
02
Write the Binomial Expansion Formula
The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\), where \(\binom{n}{k}\) are the binomial coefficients. For \( (1 + \sqrt{2})^6 \), we identify \( a = 1 \) and \( b = \sqrt{2} \), and we need to expand it for \( n = 6 \).
03
Extract Coefficients from Pascal's Triangle
Locate the coefficients for \( n = 6 \) in Pascal鈥檚 Triangle. The coefficients are \([1, 6, 15, 20, 15, 6, 1]\). These will act as scalar multipliers for each respective term in the expansion.
04
Calculate Each Term in the Expansion
Use the coefficients and the formula. The expansion is: \(1^6(\sqrt{2})^0 + 6 \, 1^5(\sqrt{2})^1 + 15 \, 1^4(\sqrt{2})^2 + 20 \, 1^3(\sqrt{2})^3 + 15 \, 1^2(\sqrt{2})^4 + 6 \, 1^1(\sqrt{2})^5 + 1 \, 1^0(\sqrt{2})^6\). Compute each individual term using these formulas:
05
Simplify Each Power
Calculate each power \(1^k = 1\) for any \(k\) and \( (\sqrt{2})^k = 2^{k/2}\). Therefore, compute each term separately:- \(1\cdot(\sqrt{2})^0 = 1\)- \(6\cdot(\sqrt{2})^1 = 6\sqrt{2}\)- \(15\cdot(\sqrt{2})^2 = 15\times2 = 30\)- \(20\cdot(\sqrt{2})^3 = 20\cdot2^{3/2} = 20\cdot2\sqrt{2} = 40\sqrt{2}\)- \(15\cdot(\sqrt{2})^4 = 15\times4 = 60\)- \(6\cdot(\sqrt{2})^5 = 6\cdot4\sqrt{2} = 24\sqrt{2}\)- \(1\cdot(\sqrt{2})^6 = 8\)
06
Combine All Terms
Add all the terms together:\(1 + 6\sqrt{2} + 30 + 40\sqrt{2} + 60 + 24\sqrt{2} + 8\)Combine like terms (constant terms and terms with \(\sqrt{2}\)):- Constant terms: \(1 + 30 + 60 + 8 = 99\)- Terms with \(\sqrt{2}\): \(6\sqrt{2} + 40\sqrt{2} + 24\sqrt{2} = 70\sqrt{2}\)Thus, the expansion results in \(99 + 70\sqrt{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Binomial Expansion
Binomial expansion is a way to express the power of a binomial鈥攁 sum of two terms, typically written as \((a + b)^n\)鈥攁s a polynomial.
This is achieved using the principles of combinations and systematic multiplication of terms.
In the context of the given exercise, the binomial to expand is \((1 + \sqrt{2})^6\).
This is achieved using the principles of combinations and systematic multiplication of terms.
In the context of the given exercise, the binomial to expand is \((1 + \sqrt{2})^6\).
- The expanded form consists of terms, each with a certain coefficient, and powers of both 1 and \(\sqrt{2}\).
- The number of terms in the expansion is equivalent to \(n+1\) where \(n\) is the power of the binomial.
The Binomial Theorem Demystified
The Binomial Theorem is a powerful tool that allows us to expand expressions raised to a power in an organized way.
The theorem tells us that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
The theorem tells us that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
- Here, \(\binom{n}{k}\) denotes the binomial coefficient, representing the number of ways to choose \(k\) items from \(n\) without concern for the order.
- This formula generates terms where each one is the product of a binomial coefficient, a power of the first term \(a\), and a power of the second term \(b\).
Grasping Binomial Coefficients
Binomial coefficients are the numbers you encounter in Pascal's Triangle and are symbolized by \(\binom{n}{k}\).
They represent the number of ways to select \(k\) elements from a set of \(n\) elements without caring about the order.
They represent the number of ways to select \(k\) elements from a set of \(n\) elements without caring about the order.
- These coefficients can be calculated as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), using factorials of numbers.
- In our context, for \((1 + \sqrt{2})^6\), the coefficients from Pascal's Triangle for the sixth row鈥擻([1, 6, 15, 20, 15, 6, 1]\)鈥攁re used directly in expansion.
