Find the general term `(r+1)^("th")` term in the expansion of
`(1+(4x)/(5))^(5//2)`

Answer

To find the general term \( (r+1)^{th} \) in the expansion of \( \left(1 + \frac{4x}{5}\right)^{\frac{5}{2}} \), we can use the Binomial Theorem, which states that:

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

In our case, we can identify \( a = 1 \), \( b = \frac{4x}{5} \), and \( n = \frac{5}{2} \).

### Step 1: Identify the general term
The general term \( T_{r+1} \) in the expansion can be expressed as:

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

Substituting the values of \( a \), \( b \), and \( n \):

\[ T_{r+1} = \binom{\frac{5}{2}}{r} \cdot 1^{\frac{5}{2}-r} \cdot \left(\frac{4x}{5}\right)^r \]

### Step 2: Simplify the term
Since \( 1^{\frac{5}{2}-r} = 1 \), we can simplify this to:

\[ T_{r+1} = \binom{\frac{5}{2}}{r} \cdot \left(\frac{4x}{5}\right)^r \]

### Step 3: Calculate the binomial coefficient
The binomial coefficient \( \binom{\frac{5}{2}}{r} \) can be calculated using the formula:

\[ \binom{n}{r} = \frac{n(n-1)(n-2)\cdots(n-r+1)}{r!} \]

For \( n = \frac{5}{2} \):

\[ \binom{\frac{5}{2}}{r} = \frac{\frac{5}{2} \left(\frac{5}{2}-1\right) \left(\frac{5}{2}-2\right) \cdots \left(\frac{5}{2}-(r-1)\right)}{r!} \]

### Step 4: Substitute back into the general term
Now substituting this back into our expression for \( T_{r+1} \):

\[ T_{r+1} = \frac{\frac{5}{2} \cdot \left(\frac{3}{2}\right) \cdots \left(\frac{5}{2} - (r-1)\right)}{r!} \cdot \left(\frac{4x}{5}\right)^r \]

### Final Expression
Thus, the general term \( T_{r+1} \) in the expansion of \( \left(1 + \frac{4x}{5}\right)^{\frac{5}{2}} \) is:

\[ T_{r+1} = \frac{\frac{5}{2} \cdot \frac{3}{2} \cdots \left(\frac{5}{2} - (r-1)\right)}{r!} \cdot \left(\frac{4x}{5}\right)^r \]