If the 21st and 22nd terms in the expansion of `(1 - x)^(44)` are equal, then x =

Options

A. `7//8`
B. `-7//8`
C. `8//7`
D. `-8//7`

Answer

To solve the problem, we need to find the value of \( x \) such that the 21st and 22nd terms in the expansion of \( (1 - x)^{44} \) are equal.

### Step-by-Step Solution:

1. **Identify the General Term in the Expansion**:
The \( r \)-th term in the binomial expansion of \( (1 - x)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} (-x)^r \]
For our case, \( n = 44 \).

2. **Write the 21st and 22nd Terms**:
- The 21st term (\( T_{21} \)) corresponds to \( r = 20 \):
\[ T_{21} = \binom{44}{20} (-x)^{20} \]
- The 22nd term (\( T_{22} \)) corresponds to \( r = 21 \):
\[ T_{22} = \binom{44}{21} (-x)^{21} \]

3. **Set the Two Terms Equal**:
Since the 21st and 22nd terms are equal, we can set them equal to each other:
\[ \binom{44}{20} (-x)^{20} = \binom{44}{21} (-x)^{21} \]

4. **Simplify the Equation**:
We can simplify the equation by dividing both sides by \( (-x)^{20} \) (assuming \( x \neq 0 \)):
\[ \binom{44}{20} = \binom{44}{21} (-x) \]

5. **Use the Property of Binomial Coefficients**:
Recall that:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Therefore, we can express \( \binom{44}{21} \) as:
\[ \binom{44}{21} = \frac{44!}{21! \cdot 23!} \]
and \( \binom{44}{20} \) as:
\[ \binom{44}{20} = \frac{44!}{20! \cdot 24!} \]

6. **Substituting the Binomial Coefficients**:
Substitute these into the equation:
\[ \frac{44!}{20! \cdot 24!} = \frac{44!}{21! \cdot 23!} (-x) \]

7. **Canceling \( 44! \)**:
Cancel \( 44! \) from both sides:
\[ \frac{1}{20! \cdot 24!} = \frac{(-x)}{21! \cdot 23!} \]

8. **Cross-Multiplying**:
Cross-multiply to eliminate the fractions:
\[ 21! \cdot 23! = -x \cdot 20! \cdot 24! \]

9. **Simplifying Further**:
We know \( 21! = 21 \cdot 20! \) and \( 23! = 23 \cdot 22 \cdot 21! \):
\[ 21 \cdot 23 \cdot 22 \cdot 20! = -x \cdot 20! \cdot 24! \]
Dividing both sides by \( 20! \):
\[ 21 \cdot 23 \cdot 22 = -x \cdot 24! \]

10. **Finding \( x \)**:
Now, we can express \( 24! \) as \( 24 \cdot 23! \):
\[ 21 \cdot 23 \cdot 22 = -x \cdot 24 \cdot 23! \]
Rearranging gives:
\[ x = -\frac{21 \cdot 23 \cdot 22}{24} \]

11. **Calculating the Final Value**:
Simplifying the expression:
\[ x = -\frac{21 \cdot 23 \cdot 22}{24} = -\frac{7 \cdot 23 \cdot 22}{8} = -\frac{7 \cdot 506}{8} = -\frac{3542}{8} = -\frac{7}{8} \]

### Final Answer:
\[ x = -\frac{7}{8} \]