Understanding Binomial Expansion and Middle Terms

The question asks about the expansion of the expression \((1 + x)^{2n + 1}\). This is a binomial expansion of the form \((a + b)^N\), where \(a = 1\), \(b = x\), and the exponent \(N = 2n + 1\). Since \(n\) is typically an integer, \(2n + 1\) is an odd integer. The total number of terms in the expansion of \((1 + x)^N\) is \(N + 1\). In this case, the number of terms is \((2n + 1) + 1 = 2n + 2\).

When the number of terms in a binomial expansion is even, there are two 'middle' terms. For an expansion \((a+b)^N\) where \(N\) is odd (total terms \(N+1\) is even), the middle terms are the \(\left(\frac{N+1}{2}\right)\)-th term and the \(\left(\frac{N+3}{2}\right)\)-th term.

In our expansion \((1 + x)^{2n + 1}\), \(N = 2n + 1\). The positions of the two middle terms are:

• First middle term: \(\frac{(2n + 1) + 1}{2} = \frac{2n + 2}{2} = (n + 1)\)-th term.
• Second middle term: \(\frac{(2n + 1) + 3}{2} = \frac{2n + 4}{2} = (n + 2)\)-th term.

So, the two middle terms are the \((n+1)\)-th term and the \((n+2)\)-th term.

Finding the Coefficients of the Middle Terms

The general term (the \((r+1)\)-th term, denoted by \(T_{r+1}\)) in the binomial expansion of \((1 + x)^N\) is given by the formula:

\(T_{r+1} = \binom{N}{r} x^r\)

The coefficient of the \((r+1)\)-th term is \(\binom{N}{r}\).

For the \((n+1)\)-th term, we have \(r+1 = n+1\), which means \(r = n\). The coefficient of the \((n+1)\)-th term is \(\binom{2n + 1}{n}\).

For the \((n+2)\)-th term, we have \(r+1 = n+2\), which means \(r = n+1\). The coefficient of the \((n+2)\)-th term is \(\binom{2n + 1}{n + 1}\).

The two middle coefficients are \(\binom{2n + 1}{n}\) and \(\binom{2n + 1}{n + 1}\).

Calculating the Average of the Coefficients

The average of these two coefficients is the sum of the coefficients divided by 2:

Average = \(\frac{\binom{2n + 1}{n} + \binom{2n + 1}{n + 1}}{2}\)

We can use the property of binomial coefficients that states \(\binom{N}{k} = \binom{N}{N - k}\). Let's apply this property to the second coefficient, \(\binom{2n + 1}{n + 1}\):

\(\binom{2n + 1}{n + 1} = \binom{2n + 1}{(2n + 1) - (n + 1)} = \binom{2n + 1}{2n + 1 - n - 1} = \binom{2n + 1}{n}\)

This shows that the two middle coefficients, \(\binom{2n + 1}{n}\) and \(\binom{2n + 1}{n + 1}\), are actually equal.

Now substitute this back into the average calculation:

Average = \(\frac{\binom{2n + 1}{n} + \binom{2n + 1}{n}}{2}\)

Average = \(\frac{2 \times \binom{2n + 1}{n}}{2}\)

Average = \(\binom{2n + 1}{n}\)

Thus, the average of the coefficients of the two middle terms in the expansion of \((1 + x)^{2n + 1}\) is \(\binom{2n + 1}{n}\).

Comparing with Options

Let's compare our result with the given options:

Our calculated average, \(\binom{2n + 1}{n}\), matches Option 2.

Revision Table: Binomial Expansion Concepts

Additional Information: Properties of Binomial Coefficients

Binomial coefficients, denoted as \(\binom{N}{k}\) or \(^N C_k\), are fundamental in combinatorics and the binomial theorem. They represent the number of ways to choose \(k\) elements from a set of \(N\) elements without regard to the order of selection. Here are some key properties:

• Definition: \(\binom{N}{k} = \frac{N!}{k!(N-k)!}\) for \(0 \le k \le N\), where \(N!\) is the factorial of \(N\).
• Symmetry Property: \(\binom{N}{k} = \binom{N}{N-k}\). This is the property we used to show the two middle coefficients were equal. It reflects that choosing \(k\) items from \(N\) is the same as choosing \(N-k\) items to leave behind.
• Pascal's Identity: \(\binom{N}{k} + \binom{N}{k+1} = \binom{N+1}{k+1}\). This identity is the basis for constructing Pascal's Triangle, where each number is the sum of the two numbers directly above it.
• Sum of Coefficients: The sum of all coefficients in the expansion of \((1+x)^N\) is \(\sum_{k=0}^{N} \binom{N}{k} = 2^N\). This can be shown by setting \(x=1\) in the expansion \((1+x)^N = \sum_{k=0}^{N} \binom{N}{k} x^k\).

Understanding these properties is crucial for solving problems involving binomial expansions and coefficients.