Problem 35 - Precalculus: Mathematical for Calculus

Question

Find the term containing \(x^{4}\) in the expansion of \((x+2 y)^{10}\)

Step-by-Step Solution

Answer

The term containing \(x^4\) is \(13440x^4y^6\).

Step 1: Understanding the Binomial Theorem

We use the binomial theorem to expand expressions of the form \((a+b)^n\). According to the binomial theorem, \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).

Step 2: Identifying Variables and Indices

In our expression \((x+2y)^{10}\), set \(a = x\) and \(b = 2y\). We want the term where the power of \(x\) is 4, which corresponds to \(k=6\) since \(n-k=4\) in the binomial term.

Step 3: Calculating the Binomial Coefficient

Substitute \(n=10\) and \(k=6\) into the binomial coefficient formula: \(\binom{n}{k} = \binom{10}{6} = \frac{10!}{6!(10-6)!} = 210\).

Step 4: Substituting Powers of Terms

Plug back \(k=6\) into the binomial term \(\binom{10}{6} x^{10-6} (2y)^6\). Simplify to get \(\binom{10}{6} x^4 (2y)^6 = 210 x^4 (2^6) y^6\).

Step 5: Simplifying the Expression

Calculate \((2^6) = 64\), thus making the term \(210 \times 64 x^4 y^6\). This simplifies to \(13440 x^4 y^6\).

Step 6: Identifying the Term

The term containing \(x^4\) is therefore \(13440 x^4 y^6\).

Key Concepts

Binomial Coefficient

The binomial coefficient is a key part of the binomial theorem, widely used for expanding powers of a binomial expression like \((a+b)^n\).
You will often encounter it in the form \( \binom{n}{k} \), pronounced "n choose k", which tells you how many combinations of \( n \) items can be picked \( k \) at a time. The formula for the binomial coefficient is \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where the exclamation mark "\(!\)" denotes factorial, meaning you multiply all integers from 1 up to that number.
In our exercise, determining \( \binom{10}{6} \) was crucial to find the term containing \( x^4 \) in the expansion. Using the binomial coefficient helps in assigning appropriate powers to \( a \) and \( b \) in the binomial expansion, leading us to the correct term.
Understanding how to calculate these coefficients is foundational when dealing with polynomial expansions using the binomial theorem.

Polynomial Expansion

Polynomial expansion using the binomial theorem simplifies calculating the expansion of expressions like \((a+b)^n\).
According to the theorem, this expansion is represented by \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]which means you sum up terms where each has a binomial coefficient \( \binom{n}{k} \), with \( a \) raised to \( n-k \) and \( b \) raised to \( k \).
In simple terms, this expansion distributes the powers of \( a \) and \( b \) across each term, adjusted by their respective coefficients. In our problem, by setting \( a = x \) and \( b = 2y \), and determining that \( n-k = 4 \), it allowed us to isolate the term where \( x \) was raised to the fourth power.
The polynomial expansion thus serves as a method to fully express binomials to any power, breaking them into simpler, individual polynomial terms, which is especially useful in combinatorics and other mathematical fields.

Combinatorics

Combinatorics is the branch of mathematics dealing with combinations and arrangements of objects.
It's a fascinating field that applies to various problems, including those that involve expanding polynomials with the binomial theorem. It provides the underlying reasoning and formulae for determining how terms are combined within polynomial expansions.

• For example, the binomial coefficient \( \binom{n}{k} \) directly taps into combinatorics by calculating the number of ways to select \( k \) elements from a set of \( n \) elements.
• This directly applies to determining how terms like \( x^4 \) arise in expansions such as \((x+2y)^{10}\).

Whether you're organizing data, optimizing processes, or expanding polynomials, combinatorial approaches often reveal elegant solutions.
Understanding these concepts help make sense of the systematic way in which large sets of possibilities are counted, arranged, and chosen, enabling mathematicians to solve complex problems in simple steps.