Problem 40 - Precalculus Mathematics for Calculus

Question

Find the term containing \(y^{3}\) in the expansion of \((\sqrt{2}+y)^{12}\).

Step-by-Step Solution

Answer

The term containing \( y^3 \) is \( 7040 \sqrt{2} y^3 \).

Step 1: Recognize the Binomial Theorem

The binomial theorem states that for any positive integer \( n \), \((a + b)^n\) can be expanded as \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this problem, \( a = \sqrt{2} \) and \( b = y \).

Step 2: Identify Term Containing \( y^3 \)

The term in the expansion \( (\sqrt{2} + y)^{12} \) which contains \( y^3 \) corresponds to the term where the power of \( y \) is 3. This occurs when \( k = 3 \) in the binomial expansion formula.

Step 3: Apply Binomial Coefficient and Powers

When \( k = 3 \), the term in the expansion is \( \binom{12}{3} (\sqrt{2})^{12-3} y^3 \). Compute each part: \( \binom{12}{3} \) is the binomial coefficient, and the power of \( \sqrt{2} \) is \( 12 - 3 = 9 \).

Step 4: Calculate Binomial Coefficient

Calculate the binomial coefficient using the formula: \( \binom{12}{3} = \frac{12!}{3!(12-3)!} = 220 \).

Step 5: Calculate the Power of \( \sqrt{2} \)

Compute \((\sqrt{2})^9\). Since \( \sqrt{2} = 2^{1/2} \), \((\sqrt{2})^9 = (2^{1/2})^9 = 2^{9/2} = 32 \sqrt{2}\).

Step 6: Formulate the Term

Substitute into the expression: \( \binom{12}{3} (\sqrt{2})^9 y^3 = 220 \times 32 \sqrt{2} \times y^3 = 7040 \sqrt{2} y^3 \).

Step 7: Conclusion

Thus, the term containing \( y^3 \) in the expansion is \( 7040 \sqrt{2} y^3 \).

Key Concepts

Binomial Coefficient

The binomial coefficient is a key part of the Binomial Theorem. It provides a way to calculate the number of ways to choose a certain number of elements from a larger set, which is essential in expanding binomials.
In this context, the binomial coefficient is represented as \(\binom{n}{k}\), where \(n\) is the total number of items, and \(k\) is the number of items being chosen. It can be calculated using the formula:

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

In the original exercise, we use the coefficient \(\binom{12}{3}\) to find the term in the expansion where the exponent of \(y\) is 3. This coefficient tells us how many different ways we can choose three \(y\)s in the binomial expansion of \((\sqrt{2} + y)^{12}\). By calculating this, we find it's equal to 220, simplifying the process of expanding the polynomial.

Polynomial Expansion

Polynomial expansion involves expressing a binomial raised to a power as a sum of terms. Each term in this sum is made up of products of the binomial's elements raised to varying powers.
Here, we utilize the Binomial Theorem to expand \((a + b)^n\) into:

• \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

For the given problem, \(a = \sqrt{2}\) and \(b = y\), creating an expansion with multiple terms. Each term's power of \(y\) increases while the power of \(\sqrt{2}\) decreases as the index \(k\) increases. The challenge is to identify the term where \(y\) specifically appears with a power of 3.
The expansion thus gives us a systematic way to find any particular power of \(y\) by selecting the relevant index \(k\).

Exponents and Radicals

Understanding exponents and radicals is crucial when dealing with elements like \(\sqrt{2}\) raised to the power. Exponents indicate how many times a base is multiplied by itself, while a radical, like a square root, represents what number, when multiplied by itself, will yield the original number.
When \(\sqrt{2}\) is involved in expressions, it is helpful to express it in terms of exponents: \(\sqrt{2} = 2^{1/2}\). For powers, such as \((\sqrt{2})^9\), this can be further simplified using exponent rules:

• \((2^{1/2})^9 = 2^{9/2}\)

This can then be interpreted as \(2^4 \times \sqrt{2} = 16 \times \sqrt{2} = 32\sqrt{2}\). This understanding is key to simplifying polynomial terms and ensuring each term is correctly calculated during expansion as seen in our exercise.