Problem 15
Question
Use Pascal’s triangle to expand the expression. $$ \left(\frac{1}{x}-\sqrt{x}\right)^{5} $$
Step-by-Step Solution
Verified Answer
Expanded expression: \( \frac{1}{x^5} - \frac{5\sqrt{x}}{x^4} + \frac{10}{x^3} - \frac{10\sqrt{x}}{x^2} + \frac{5}{x} - x^{5/2} \).
1Step 1: Understand the Expression
The expression to expand is \( \left( \frac{1}{x} - \sqrt{x} \right)^5 \). This means we need to find the coefficients and powers for each term in the expansion using the binomial theorem.
2Step 2: Use the Binomial Theorem
The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). Here, \(a = \frac{1}{x}\), \(b = -\sqrt{x}\), and \(n = 5\).
3Step 3: Identify Coefficients Using Pascal's Triangle
The coefficients for the expansion are given by the 6th row of Pascal’s triangle: 1, 5, 10, 10, 5, 1.
4Step 4: Calculate Each Term
For each term, use the formula: \( \binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (-\sqrt{x})^k \).- For \(k = 0\): \( \binom{5}{0} \left( \frac{1}{x} \right)^5 = \frac{1}{x^5} \)- For \(k = 1\): \( \binom{5}{1} \left( \frac{1}{x} \right)^4 (-\sqrt{x}) = -\frac{5}{x^4}\sqrt{x} \)- For \(k = 2\): \( \binom{5}{2} \left( \frac{1}{x} \right)^3 (\sqrt{x})^2 = \frac{10}{x^3}\)- For \(k = 3\): \( \binom{5}{3} \left( \frac{1}{x} \right)^2 (-\sqrt{x})^3 = -\frac{10}{x^2}\sqrt{x}\)- For \(k = 4\): \( \binom{5}{4} \left( \frac{1}{x} \right)^1 (\sqrt{x})^4 = \frac{5}{x}\)- For \(k = 5\): \( \binom{5}{5} (\sqrt{x})^5 = -x^{5/2} \)
5Step 5: Combine All Terms
Combine all the individual terms derived from Step 4 to write the expanded expression:\[ \frac{1}{x^5} - \frac{5\sqrt{x}}{x^4} + \frac{10}{x^3} - \frac{10\sqrt{x}}{x^2} + \frac{5}{x} - x^{5/2}. \]
Key Concepts
Binomial TheoremExpression ExpansionPascal's Triangle Coefficients
Binomial Theorem
The Binomial Theorem is a fundamental principle in algebra that provides a way to expand expressions that are raised to a power. It can be written as \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this formula:
- \(a\) and \(b\) are any real or complex numbers or expressions.
- \(n\) is a non-negative integer indicating the power to which the binomial is raised.
- The sigma notation \(\sum\) indicates that we sum all terms from \(k = 0\) to \(k = n\).
- \(\binom{n}{k}\) represents the binomial coefficient, which can be calculated as \(\frac{n!}{k!(n-k)!}\).
Expression Expansion
Expression expansion involves expressing a power of a binomial as a sum of individual terms. Using the Binomial Theorem, each term in the expansion of \(\left(\frac{1}{x} - \sqrt{x}\right)^5\) can be calculated systematically:
- Determine the terms from the expansion formula \(\binom{n}{k} a^{n-k} b^k\).
- For each term, identify the power of each component: \((\frac{1}{x})^{5-k}\) and \((-\sqrt{x})^k\).
- Multiply by the coefficient from Pascal's Triangle.
Pascal's Triangle Coefficients
Pascal's Triangle is a triangular array of numbers, where each number is the sum of the two directly above it. It's an efficient way to find the binomial coefficients \(\binom{n}{k}\) needed for binomial expansion. Each row corresponds to an exponent \(n\) in the binomial expansion.
To expand \(\left(\frac{1}{x} - \sqrt{x}\right)^5\), we refer to the 6th row of Pascal's Triangle, which contains the coefficients: 1, 5, 10, 10, 5, 1. These coefficients:
To expand \(\left(\frac{1}{x} - \sqrt{x}\right)^5\), we refer to the 6th row of Pascal's Triangle, which contains the coefficients: 1, 5, 10, 10, 5, 1. These coefficients:
- Dictate the magnitude of each term in the expanded expression.
- Help simplify expression handling, avoiding tedious calculations for coefficients each time.
Other exercises in this chapter
Problem 15
\(13-20\) . Determine whether the sequence is arithmetic. If is arithmetic, find the common difference. $$ 2,4,8,16, \dots $$
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\(13-18\) Find the first five terms of the given recursively defined sequence. $$ a_{n}=2 a_{n-1}+1 \quad \text { and } \quad a_{1}=1 $$
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Show that \(n^{2}+n\) is divisible by 2 for all natural numbers \(n\)
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Determine whether the sequence is geometric. If is geometric, find the common ratio. $$ 3, \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \dots $$
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