Problem 64 - Precalculus with Limits

Question

In Exercises 61 - 66, use the Binomial Theorem to expand and simplify the expression. \( \left(u^{3/5} + 2\right)^5 \)

Step-by-Step Solution

Answer

The expanded form of the expression \((u^{3/5} + 2)^5\) using the Binomial theorem is \(u^3 + 5u^{12/5} + 40u^{9/5} + 160u^{6/5} + 320u^{3/5} + 32\).

Step 1: Understand the Binomial Theorem

The Binomial Theorem states that for any numbers \(x\) and \(y\), and any natural number \(n\), the expansion of \((x + y)^n\) is given by \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\), where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) denotes the binomial coefficient, and '!' is the factorial symbol.

Step 2: Identify the variables in the expression

Here, \(x\) corresponds to \(u^{3/5}\), \(y\) corresponds to \(2\), and \(n = 5\). This is the format needed to directly apply the Binomial Theorem.

Step 3: Apply the Binomial Theorem

Using the formula, the expansion of \((u^{3/5} + 2)^5\) is \(\sum_{k=0}^{5} \binom{5}{k} (u^{3/5})^{5-k} * 2^k\).

Step 4: Compute the Terms of the Expansion

Calculate each term of the sum separately: \binom{5}{0} (u^n)^5 * 2^0 = u^3 , \binom{5}{1} (u^n)^4 * 2^1 = 5u^{12/5}, \binom{5}{2} (u^n)^3 * 2^2 = 40u^{9/5}, \binom{5}{3} (u^n)^2 * 2^3 = 160u^{6/5}, \binom{5}{4} (u^n) * 2^4 = 320u^{3/5}, and \binom{5}{5} * 2^5 = 32.

Step 5: Write the final expression

Combine all terms to write the final, expanded expression: \(u^3 + 5u^{12/5} + 40u^{9/5} + 160u^{6/5} + 320u^{3/5} + 32\).

Key Concepts

Binomial Expansion

When we talk about binomial expansion, we're referring to the process of expanding expressions like \((x + y)^n\) using the Binomial Theorem. This allows us to express the power of a binomial as a sum of terms involving coefficients, variables, and exponents.
For example, in the expression \((u^{3/5} + 2)^5\), we identify \(x\) as \(u^{3/5}\) and \(y\) as \(2\). Using the Binomial Theorem, we write it as:

• \(\sum_{k=0}^{5} \binom{5}{k} (u^{3/5})^{5-k} \cdot 2^k\)

Each term in the expansion is derived from adjusting the powers of \(x\) and \(y\), and is multiplied by a binomial coefficient.
This organized approach allows polynomial expressions to be expanded systematically, making them easier to handle, especially with higher or fractional exponents.

Binomial Coefficients

The binomial coefficients are critical components in the Binomial Theorem. They are represented as \(\binom{n}{k}\), also read as "n choose k". These coefficients determine how many ways we can pick \(k\) items from \(n\) items without regard to the order.
The formula for calculating a binomial coefficient is:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

In our example of \((u^{3/5} + 2)^5\), the coefficients \(\binom{5}{k}\) are used for each term:

• \(\binom{5}{0} = 1\)
• \(\binom{5}{1} = 5\)
• \(\binom{5}{2} = 10\)
• \(\binom{5}{3} = 10\)
• \(\binom{5}{4} = 5\)
• \(\binom{5}{5} = 1\)

These coefficients help determine the multiplicative factor for each term in the expansion. Understanding how these work is essential for accurately expanding a binomial expression.

Factorials

Factorials are a crucial part of calculating binomial coefficients, represented by the exclamation mark (!). Given a non-negative integer \(n\), the factorial \(n!\) is the product of all positive integers less than or equal to \(n\).
For instance:

• \(0! = 1\)
• \(1! = 1\)
• \(2! = 2 \cdot 1 = 2\)
• \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\)

Factorials are used in the computation of binomial coefficients because each binomial coefficient requires the calculation of three factorials: \(n!\), \(k!\), and \((n-k)!\).
For example, \(\binom{5}{2}\):

• Calculate \(5! = 120\)
• Calculate \(2! = 2\)
• Calculate \(3! = 6\) (since \(5-2=3\))
• Finally, \(\binom{5}{2} = \frac{120}{2 \times 6} = 10\)

Having a strong grasp of factorials is essential not just in binomial expansion, but in various fields of mathematics.