Chapter 11: Problem 5
Use the binomial theorem to expand. a) \((x+y)^{2}\) b) \((a+1)^{3}\) c) \((1-p)^{4}\)
Short Answer
Expert verified
a) \(x^2 + 2xy + y^2\), b) \(a^3 + 3a^2 + 3a + 1\), c) \(1 - 4p + 6p^2 - 4p^3 + p^4\).
Step by step solution
01
Title - Understand the Binomial Theorem
The Binomial Theorem states that \ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k \ where \(\binom{n}{k}\) is the binomial coefficient calculated as \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
02
Title - Expand \ (x + y)^2
For \((x + y)^2\), we have \(n = 2\). The expansion is: \ (x + y)^2 = \binom{2}{0} x^2 y^0 + \binom{2}{1} x^1 y^1 + \binom{2}{2} x^0 y^2\ \ Calculating the coefficients, we get: \ \binom{2}{0} = 1, \binom{2}{1} = 2, \binom{2}{2} = 1\ \ Therefore, \( (x+y)^{2} = 1x^2 + 2xy + 1y^2 = x^2 + 2xy + y^2 \)
03
Title - Expand \ (a + 1)^3
For \((a + 1)^3\), we have \(n = 3\). The expansion is: \ (a + 1)^3 = \binom{3}{0} a^3 1^0 + \binom{3}{1} a^2 1^1 + \binom{3}{2} a^1 1^2 + \binom{3}{3} a^0 1^3\ \ Calculating the coefficients, we get: \ \binom{3}{0} = 1, \binom{3}{1} = 3, \binom{3}{2} = 3, \binom{3}{3} = 1\ \ Therefore, \( (a+1)^{3} = 1a^3 + 3a^2 + 3a + 1 = a^3 + 3a^2 + 3a + 1 \)
04
Title - Expand \ (1 - p)^4
For \((1 - p)^4\), we have \(n = 4\). The expansion is: \ (1 - p)^4 = \binom{4}{0} 1^4 (-p)^0 + \binom{4}{1} 1^3 (-p)^1 + \binom{4}{2} 1^2 (-p)^2 + \binom{4}{3} 1^1 (-p)^3 + \binom{4}{4} 1^0 (-p)^4\ \ Calculating the coefficients, we get: \ \binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4, \binom{4}{4} = 1\ \ Therefore, \( (1-p)^{4} = 1 - 4p + 6p^2 - 4p^3 + p^4 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Expansion
The binomial theorem provides a straightforward way to expand expressions that are raised to a power. Essentially, binomial expansion covers breaking down an expression with two terms, such as \(x + y\), into a sum of several terms with different powers.
For instance, if you have \((x + y)^2\), the binomial theorem helps us expand it into \(x^2 + 2xy + y^2\). Neat, right? The number of terms in the expansion corresponds to the power plus one. So if you're expanding \((a + 1)^3\), you'll end up with four terms.
The binomial expansion formula is stated as follows:\[ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\]
Here, \(\binom{n}{k}\) is the binomial coefficient. Don’t worry, we’ll break it down further in the next section.
For instance, if you have \((x + y)^2\), the binomial theorem helps us expand it into \(x^2 + 2xy + y^2\). Neat, right? The number of terms in the expansion corresponds to the power plus one. So if you're expanding \((a + 1)^3\), you'll end up with four terms.
The binomial expansion formula is stated as follows:\[ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\]
Here, \(\binom{n}{k}\) is the binomial coefficient. Don’t worry, we’ll break it down further in the next section.
Binomial Coefficients
Binomial coefficients are a crucial part of the binomial theorem. They determine the weights of each term in the expansion. These coefficients are represented as \(\binom{n}{k}\) and can be calculated using the combinations formula:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
The exclamation mark here denotes a factorial. For example, \(4! = 4 \times 3 \times 2 \times 1\).
Let’s say we’re dealing with \((1 - p)^4\). We calculate the binomial coefficients for every term using the formula. So, \(\binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4,\ and \binom{4}{4} = 1\). If you apply these coefficients to \((1 - p)^4\), you'll get the expansion: \(1 - 4p + 6p^2 - 4p^3 + p^4\).
In each term of the expansion, the coefficients multiply the respective terms to give us the final expanded form.
The exclamation mark here denotes a factorial. For example, \(4! = 4 \times 3 \times 2 \times 1\).
Let’s say we’re dealing with \((1 - p)^4\). We calculate the binomial coefficients for every term using the formula. So, \(\binom{4}{0} = 1, \binom{4}{1} = 4, \binom{4}{2} = 6, \binom{4}{3} = 4,\ and \binom{4}{4} = 1\). If you apply these coefficients to \((1 - p)^4\), you'll get the expansion: \(1 - 4p + 6p^2 - 4p^3 + p^4\).
In each term of the expansion, the coefficients multiply the respective terms to give us the final expanded form.
Combinations
Combinations play a huge role in figuring out binomial coefficients. In general, a combination is a way of selecting items from a larger pool, where the order does not matter. The formula for combinations is the same as for binomial coefficients:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
For instance, if you want to select 2 items out of 5, the number of ways you can do that is given by \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \).
Now that we know how to compute combinations, we use them straight in the binomial formula to get our coefficients. Let’s revisit a previous example: \((a + 1)^3\). Here, we compute \(\binom{3}{0}, \binom{3}{1}, \binom{3}{2},\ and \binom{3}{3}\). Using our combinations knowledge, we'll know these are 1, 3, 3, and 1. With these coefficients, we can write the expanded form: \(a^3 + 3a^2 + 3a + 1\). Understanding combinations essentially unlocks the door to mastering binomial expansions!
For instance, if you want to select 2 items out of 5, the number of ways you can do that is given by \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = 10 \).
Now that we know how to compute combinations, we use them straight in the binomial formula to get our coefficients. Let’s revisit a previous example: \((a + 1)^3\). Here, we compute \(\binom{3}{0}, \binom{3}{1}, \binom{3}{2},\ and \binom{3}{3}\). Using our combinations knowledge, we'll know these are 1, 3, 3, and 1. With these coefficients, we can write the expanded form: \(a^3 + 3a^2 + 3a + 1\). Understanding combinations essentially unlocks the door to mastering binomial expansions!