Find the
`6^("th")` term of `(1+(x)/(2))^(-5)`

Answer

To find the sixth term of the expression \( (1 + \frac{x}{2})^{-5} \) using the Binomial Theorem, we can follow these steps:

### Step 1: Identify the parameters for the Binomial Theorem
The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
In our case, \( a = 1 \), \( b = \frac{x}{2} \), and \( n = -5 \).

### Step 2: Write the formula for the \( k \)-th term
The \( k \)-th term (where \( k \) starts from 0) in the expansion is given by:
\[ T_{k+1} = \binom{n}{k} a^{n-k} b^k \]
For our expression, this becomes:
\[ T_{k+1} = \binom{-5}{k} (1)^{-5-k} \left(\frac{x}{2}\right)^k \]
This simplifies to:
\[ T_{k+1} = \binom{-5}{k} \left(\frac{x}{2}\right)^k \]

### Step 3: Find the sixth term
To find the sixth term, we need \( k = 5 \) (since \( k \) starts from 0). Thus, we need to calculate \( T_6 \):
\[ T_6 = \binom{-5}{5} \left(\frac{x}{2}\right)^5 \]

### Step 4: Calculate \( \binom{-5}{5} \)
Using the formula for binomial coefficients with negative integers:
\[ \binom{-n}{r} = (-1)^r \binom{n+r-1}{r} \]
For \( n = 5 \) and \( r = 5 \):
\[ \binom{-5}{5} = (-1)^5 \binom{5+5-1}{5} = -\binom{9}{5} \]
Calculating \( \binom{9}{5} \):
\[ \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]
Thus, \( \binom{-5}{5} = -126 \).

### Step 5: Calculate \( \left(\frac{x}{2}\right)^5 \)
Now we calculate:
\[ \left(\frac{x}{2}\right)^5 = \frac{x^5}{2^5} = \frac{x^5}{32} \]

### Step 6: Combine the results
Now, substituting back into the term:
\[ T_6 = -126 \cdot \frac{x^5}{32} = -\frac{126x^5}{32} \]

### Final Answer
The sixth term of the expression \( (1 + \frac{x}{2})^{-5} \) is:
\[ -\frac{126x^5}{32} \]