Understanding the Problem: Binomial Expansion and Middle Terms

The question asks us to find a relationship between the coefficients of the middle term(s) of two different binomial expansions: $(1+x)^{2n}$ and $(1+x)^{2n-1}$. We are given specific names for these coefficients: $\alpha$ for the middle term of $(1+x)^{2n}$, and $\beta$ and $\gamma$ for the two middle terms of $(1+x)^{2n-1}$.

Let's first understand what a middle term is in a binomial expansion $(a+b)^N$. The expansion has $N+1$ terms.

• If $N$ is even, there is one middle term. The number of terms is $N+1$ (odd). The middle term is the $(\frac{N+1+1}{2})$-th term, which is the $(\frac{N}{2}+1)$-th term.
• If $N$ is odd, there are two middle terms. The number of terms is $N+1$ (even). The middle terms are the $(\frac{N+1}{2})$-th term and the $(\frac{N+1}{2}+1)$-th term.

The general term (T$_{k+1}$) in the expansion of $(1+x)^N$ is given by the formula $T_{k+1} = \binom{N}{k} x^k$. The coefficient of this term is $\binom{N}{k}$.

Calculating the Coefficient $\alpha$

The first expansion is $(1+x)^{2n}$. Here, $N = 2n$. Since $2n$ is an even number, the number of terms is $2n+1$, which is odd. There is one middle term.

The position of the middle term is $(\frac{2n}{2}+1)$-th term, which is the $(n+1)$-th term.

The coefficient of the $(n+1)$-th term (T$_{n+1}$) in $(1+x)^{2n}$ is $\binom{2n}{n}$.

According to the problem, this coefficient is $\alpha$. So, $\alpha = \binom{2n}{n}$.

Calculating the Coefficients $\beta$ and $\gamma$

The second expansion is $(1+x)^{2n-1}$. Here, $N = 2n-1$. Since $2n-1$ is an odd number (assuming $n \ge 1$), the number of terms is $(2n-1)+1 = 2n$, which is even. There are two middle terms.

The positions of the two middle terms are:

• The $(\frac{(2n-1)+1}{2})$-th term, which is the $(\frac{2n}{2})$-th term or the $n$-th term.
• The $(\frac{(2n-1)+1}{2}+1)$-th term, which is the $(\frac{2n}{2}+1)$-th term or the $(n+1)$-th term.

The coefficient of the $n$-th term (T$_{n}$) in $(1+x)^{2n-1}$ is $\binom{2n-1}{n-1}$ (since $T_{k+1}$ has coefficient $\binom{N}{k}$, the $n$-th term $T_{(n-1)+1}$ has $k=n-1$).

The coefficient of the $(n+1)$-th term (T$_{n+1}$) in $(1+x)^{2n-1}$ is $\binom{2n-1}{n}$ (since $T_{k+1}$ has coefficient $\binom{N}{k}$, the $(n+1)$-th term $T_{n+1}$ has $k=n$).

According to the problem, the coefficients of the two middle terms are $\beta$ and $\gamma$. It doesn't explicitly state which one is which, but the options suggest a simple sum relation. Let's assign them in the order they appear (n-th term then (n+1)-th term):

So, $\beta = \binom{2n-1}{n-1}$ and $\gamma = \binom{2n-1}{n}$.

Finding the Relation between $\alpha$, $\beta$, and $\gamma$

We have the coefficients:

• $\alpha = \binom{2n}{n}$
• $\beta = \binom{2n-1}{n-1}$
• $\gamma = \binom{2n-1}{n}$

We need to find a relationship between $\binom{2n}{n}$, $\binom{2n-1}{n-1}$, and $\binom{2n-1}{n}$.

Let's recall Pascal's identity for binomial coefficients, which states that $\binom{N}{K} = \binom{N-1}{K-1} + \binom{N-1}{K}$.

Let's apply this identity with $N=2n$ and $K=n$.

$\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$

Comparing this with our expressions for $\alpha$, $\beta$, and $\gamma$, we see that:

$\alpha = \beta + \gamma$

Conclusion

Based on the definitions of $\alpha$, $\beta$, and $\gamma$ from the binomial expansions and the application of Pascal's identity, the relationship between the coefficients is $\alpha = \beta + \gamma$.

Using Pascal's identity $\binom{N}{K} = \binom{N-1}{K-1} + \binom{N-1}{K}$ with $N=2n$ and $K=n$, we get $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$, which translates to $\alpha = \beta + \gamma$.

Revision Table: Key Binomial Concepts

Additional Information: Pascal's Triangle and Pascal's Identity

Pascal's identity $\binom{N}{K} = \binom{N-1}{K-1} + \binom{N-1}{K}$ is the rule that generates Pascal's triangle. Each number in Pascal's triangle is the sum of the two numbers directly above it.

The rows of Pascal's triangle correspond to the binomial coefficients $\binom{N}{k}$ for increasing values of $N$.

• Row 0 (N=0): $\binom{0}{0}$ (1)
• Row 1 (N=1): $\binom{1}{0}, \binom{1}{1}$ (1, 1)
• Row 2 (N=2): $\binom{2}{0}, \binom{2}{1}, \binom{2}{2}$ (1, 2, 1)
• Row 3 (N=3): $\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}$ (1, 3, 3, 1)

Pascal's identity relates an entry in one row (say, row N, column K) to entries in the row above it (row N-1, columns K-1 and K). Our problem uses this identity where $\binom{2n}{n}$ (an entry in row $2n$) is the sum of $\binom{2n-1}{n-1}$ and $\binom{2n-1}{n}$ (two entries in row $2n-1$). This identity is crucial for understanding the relationships between binomial coefficients.