The `(m + 1)^(th)` term of `(x/y+y/x)^(2m + 1 )` is

Options

A. independent of x
B. a constant
C. depends on the ratio `x//y` and m
D. none of these

Answer

To find the \((m + 1)^{th}\) term of the expression \((\frac{x}{y} + \frac{y}{x})^{2m + 1}\), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, \(a = \frac{x}{y}\), \(b = \frac{y}{x}\), and \(n = 2m + 1\).

### Step 1: Identify the general term
The general term \(T_{r+1}\) in the expansion is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

Substituting \(n = 2m + 1\), \(a = \frac{x}{y}\), and \(b = \frac{y}{x}\):

\[ T_{r+1} = \binom{2m + 1}{r} \left(\frac{x}{y}\right)^{(2m + 1) - r} \left(\frac{y}{x}\right)^r \]

### Step 2: Find the \((m + 1)^{th}\) term
To find the \((m + 1)^{th}\) term, we set \(r = m\):

\[ T_{m + 1} = \binom{2m + 1}{m} \left(\frac{x}{y}\right)^{(2m + 1) - m} \left(\frac{y}{x}\right)^m \]

### Step 3: Simplify the expression
Now, simplifying \(T_{m + 1}\):

\[ T_{m + 1} = \binom{2m + 1}{m} \left(\frac{x}{y}\right)^{m + 1} \left(\frac{y}{x}\right)^m \]

This can be rewritten as:

\[ T_{m + 1} = \binom{2m + 1}{m} \cdot \frac{x^{m + 1}}{y^{m + 1}} \cdot \frac{y^m}{x^m} \]

### Step 4: Combine the terms
Now, we can combine the terms:

\[ T_{m + 1} = \binom{2m + 1}{m} \cdot \frac{x^{m + 1} \cdot y^m}{y^{m + 1} \cdot x^m} \]

This simplifies to:

\[ T_{m + 1} = \binom{2m + 1}{m} \cdot \frac{y}{x} \]

### Final Result
Thus, the \((m + 1)^{th}\) term of \((\frac{x}{y} + \frac{y}{x})^{2m + 1}\) is:

\[ T_{m + 1} = \binom{2m + 1}{m} \cdot \frac{y}{x} \]