Problem 46 - A Graphical Approach to Precalculus with Limits

Question

Work each of the following. Find the two middle terms of \(\left(-2 m^{-1}+3 n^{-2}\right)^{11}\)

Step-by-Step Solution

Answer

The two middle terms are \( 7197936m^{-6}n^{-10} \) and \( -10705920m^{-5}n^{-12} \).

Step 1: Understanding the Binomial Theorem

The expression \((-2 m^{-1} + 3 n^{-2})^{11}\) is a binomial raised to the power of 11. To find specific terms, we can use the Binomial Theorem, which states that \((a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k\).

Step 2: Determine Terms to Find

Since we're asked to find the two middle terms in a binomial expansion of degree 11, there will be a total of 12 terms (0 to 11). For an even number of terms, the two middle terms are the 6th and 7th terms (when considering 0-based indexing, these correspond to \(k=5\) and \(k=6\) when we count from 0).

Step 3: Find the 6th Term (k=5)

Using the formula for a given term in the Binomial expansion, \( T_{k+1} = \binom{n}{k} a^{n-k} b^k \), we substitute \( n = 11 \), \( a = -2m^{-1} \), \( b = 3n^{-2} \), and \( k = 5 \):\[ T_6 = \binom{11}{5} (-2m^{-1})^{6} (3n^{-2})^{5} = 462 \times 64m^{-6} \times 243n^{-10} = 7197936m^{-6}n^{-10}.\]

Step 4: Find the 7th Term (k=6)

Similarly, for \(k=6\), we find the 7th term: \[ T_7 = \binom{11}{6} (-2m^{-1})^{5} (3n^{-2})^{6} = 462 \times (-32)m^{-5} \times 729n^{-12} = -10705920m^{-5}n^{-12}.\]

Step 5: Conclusion

The two middle terms of \((-2 m^{-1}+3 n^{-2})^{11}\) are: \( 7197936m^{-6}n^{-10} \) and \( -10705920m^{-5}n^{-12} \).

Key Concepts

Binomial Expansion

The Binomial Expansion is a way to express the power of a binomial, which is an algebraic expression that consists of two terms. When you have an expression like \((a + b)^n\), the Binomial Expansion becomes extremely useful. It allows you to expand this expression into a sum of terms, each of which has a coefficient determined by a specific formula.

To expand \((a + b)^n\), the theorem states that:

• Each term in the expansion can be calculated using the general formula \( \binom{n}{k} a^{n-k} b^k \).
• Here, \( \binom{n}{k} \) is the binomial coefficient, which you can find using combinations.
• Throughout the expansion, the values of \(k\) range from 0 up to \(n\) as they explore every term.

The binomial coefficient \( \binom{n}{k} \) can be expressed as \( \frac{n!}{k!(n-k)!} \), where '!' denotes factorial, the product of all positive integers up to a specified number. This approach helps in calculating specific terms without fully expanding the expression, saving time especially with large exponents.

Middle Terms

When dealing with binomial expansions, finding the middle term(s) means identifying the term(s) that sit right in the center of the expanded expression. The total number of terms in the expansion of \((a + b)^n\) is \(n+1\).

Here's how to find the middle terms:

• If \(n\) is even, there are two middle terms. You will find these at positions \(k = \frac{n}{2} - 1\) and \(k = \frac{n}{2}\) (considering zero-based indexing).
• In terms of a practical example, if you are working with \((a + b)^{11}\), it presents an even case (12 terms). Hence, the middle terms will be the 6th term \(T_6\) and the 7th term \(T_7\).
• For \(a + b\) to efficiently get the middle terms without calculating all terms, focus just on these middle positions.

This focused method significantly reduces the work required, especially with high-degree polynomials, and allows us to find exactly what we need for analysis or calculation.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and finding patterns. It plays a critical role in the Binomial Theorem through the calculation of binomial coefficients. These coefficients are nothing but combinations (often expressed as \( \binom{n}{k} \)).

Key concepts of combinatorics include:

• Factorials (\(n!\)), which are foundational in determining permutations and combinations.
• Permutations, which are sequences where order matters, calculated as a set of arranged items.
• Combinations, which are selections of items where the order does not matter, and are defined by \( \binom{n}{k} = \frac{n!}{k!(n-k)!}\).

When addressing binomial expansions, understanding combinatorics helps to compute these values efficiently. By figuring out binomial coefficients, combinatorics provides a way to precisely expand expressions without unnecessary calculations.