Problem 36 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The term is \(1760\sqrt{2}y^{3}\).

Step 1: Identify the Binomial Theorem

To find the term in the expansion of \((\sqrt{2}+y)^{12}\), we utilize the Binomial Theorem, which states that \((a+b)^{n}\) expands to \(abla_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k}\).

Step 2: Apply the Binomial Theorem to the Expression

For \((\sqrt{2}+y)^{12}\), we assign \(a = \sqrt{2}\), \(b = y\), and \(n = 12\). The general term in this expansion is \(abla_{k=0}^{12} \binom{12}{k} (\sqrt{2})^{12-k} y^{k}\).

Step 3: Determine the Target Term Condition

We want the term where \(y^{k} = y^{3}\). Hence, we set \(k = 3\).

Step 4: Calculate the Binomial Coefficient and Powers

With \(k = 3\), calculate \( \binom{12}{3} (\sqrt{2})^{12-3} y^{3}\) which simplifies to \( \binom{12}{3} (\sqrt{2})^{9} y^{3}\).

Step 5: Compute the Binomial Coefficient

Compute \( \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220\).

Step 6: Evaluate the Power of \(\sqrt{2}\)

Calculate \( (\sqrt{2})^{9} = (2^{1/2})^{9} = 2^{9/2} = 8 \sqrt{2}\).

Step 7: Combine All Parts to Find Target Term

The term is \( 220 \times 8\sqrt{2} \times y^{3} = 1760\sqrt{2}y^{3}\).

Key Concepts

Binomial Coefficient

The binomial coefficient is a crucial element in the Binomial Theorem. It is represented by the symbol \( \binom{n}{k} \), also known as "n choose k." This tells us how many ways we can choose \( k \) elements from a set of \( n \) elements, without regard to the order. Mathematically, it is calculated as: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] where \( n! \) (n factorial) is the product of all positive integers up to \( n \).
Binomial coefficients are fundamental in determining each term of a binomial expansion by setting the coefficient of that term.

• For example, in the given problem, when \( k = 3 \) and \( n = 12 \), the binomial coefficient is \( \binom{12}{3} \), which is calculated as \( 220 \).
• This value is then used to find the coefficient of the term in the expansion.

Binomial Expansion

Binomial expansion is the process of expanding expressions that are raised to a power, like \((a + b)^n\). This expansion allows us to express the polynomial in terms of sums of powers of its terms, using coefficients determined by binomial coefficients.
The formula for binomial expansion is given by:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]This means that each term of the expanded polynomial is found by multiplying a binomial coefficient \( \binom{n}{k} \), \( a \) raised to the power of \( n-k \) and \( b \) raised to \( k \).

• In our exercise, \( a = \sqrt{2} \) and \( b = y \).
• For the expansion of \((\sqrt{2} + y)^{12}\), we used this principle to determine each term.

Powers of Roots

When working with terms like \( (\sqrt{2})^{9} \) in the expansion, we deal with powers of roots.
A root raised to a power follows the laws of exponents, where, for instance, \((a^{1/2})^b\) can be simplified to \( a^{b/2} \). So, for a term like \( (\sqrt{2})^9 \), this becomes:\[(2^{1/2})^9 = 2^{9/2}\]We can further simplify \( 2^{9/2} \) by understanding it as \( (2^{4}) \cdot \sqrt{2} = 8 \sqrt{2} \).

• This simplification is key in calculating the contribution of the root term in each part of the expanded binomial expression.
• Understanding how to manipulate such powers ensures you can accurately calculate each element in a problem like this.