Understanding the Binomial Expansion Problem

The question asks us to find a specific term in the binomial expansion of \( \left(x + \frac{1}{x}\right)^{2n} \). We need to find the \((n+1)\)th term when counting from the end of the expansion, assuming the terms are arranged in descending powers of \(x\).

The given expression is in the form of \( (a+b)^N \), where \( a = x \), \( b = \frac{1}{x} \), and \( N = 2n \).

General Term in Binomial Expansion

The general term, or the \((r+1)\)th term from the beginning, in the binomial expansion of \( (a+b)^N \) is given by the formula:

\( T_{r+1} = \binom{N}{r} a^{N-r} b^r \)

Applying the General Term Formula to \( \left(x + \frac{1}{x}\right)^{2n} \)

For the expansion \( \left(x + \frac{1}{x}\right)^{2n} \), we substitute \( a=x \), \( b=\frac{1}{x} \), and \( N=2n \) into the general term formula:

\( T_{r+1} = \binom{2n}{r} x^{2n-r} \left(\frac{1}{x}\right)^r \)

Since \( \left(\frac{1}{x}\right)^r = x^{-r} \), the formula becomes:

\( T_{r+1} = \binom{2n}{r} x^{2n-r} x^{-r} \)

Combining the powers of \(x\):

\( T_{r+1} = \binom{2n}{r} x^{2n-r-r} \)

\( T_{r+1} = \binom{2n}{r} x^{2n-2r} \)

This formula gives the \((r+1)\)th term from the beginning of the expansion.

Finding the Position of the Term from the End

The expansion of \( (a+b)^N \) has a total of \( N+1 \) terms. In our case, \( N=2n \), so the total number of terms is \( 2n+1 \).

We need to find the \((n+1)\)th term from the end.

In a series of \( M \) terms, the \( k \)th term from the end is the \((M - k + 1)\)th term from the beginning.

Here, the total number of terms \( M = 2n+1 \) and we want the \( k = n+1 \)th term from the end.

The position from the beginning is \( (2n+1 - (n+1) + 1) \).

Let's simplify the expression:

\( (2n+1 - n - 1 + 1) = (2n - n + 1 - 1 + 1) = n + 1 \)

So, the \((n+1)\)th term from the end is the same as the \((n+1)\)th term from the beginning.

Calculating the \((n+1)\)th Term from the Beginning

We use the general term formula \( T_{r+1} = \binom{2n}{r} x^{2n-2r} \).

To find the \((n+1)\)th term, we set \( r+1 = n+1 \), which means \( r=n \).

Substitute \( r=n \) into the general term formula:

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

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

\( T_{n+1} = \binom{2n}{n} x^0 \)

Since \( x^0 = 1 \), the term simplifies to:

\( T_{n+1} = \binom{2n}{n} \cdot 1 \)

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

The binomial coefficient \( \binom{2n}{n} \) is also commonly written as C(2n, n).

Comparing with the Options

We found that the \((n+1)\)th term from the end is C(2n, n). Let's check the given options:

• Option 1: C(2n, n)x
• Option 2: C(2n, n - 1)x
• Option 3: C(2n, n)
• Option 4: C(2n, n - 1)

Our result, C(2n, n), matches Option 3.

Conclusion

The \((n+1)\)th term from the end in the expansion of \( \left(x + \frac{1}{x}\right)^{2n} \), when arranged in descending powers of \(x\), is C(2n, n).

Revision Table: Key Concepts for Binomial Expansion Terms

Additional Information: Understanding the Middle Term

In the expansion of \( (a+b)^{2n} \), which has \( 2n+1 \) terms (an odd number of terms), there is a single middle term. The position of the middle term is \(\frac{2n+1+1}{2} = \frac{2n+2}{2} = n+1\).

Thus, the \((n+1)\)th term is the middle term. For the expansion \( \left(x + \frac{1}{x}\right)^{2n} \), the general term is \( T_{r+1} = \binom{2n}{r} x^{2n-2r} \). The middle term, \( T_{n+1} \), occurs when \( r=n \), resulting in \( \binom{2n}{n} x^{2n-2n} = \binom{2n}{n} x^0 = \binom{2n}{n} \).

We found that the \((n+1)\)th term from the end is also the \((n+1)\)th term from the beginning, which is the middle term. This makes sense because in an expansion with \( 2n+1 \) terms, the \((n+1)\)th term is exactly in the middle.