Find `n and x` in the expansion of `(1 + x)^n`, if the fifth term is four times the fourth term and the fourth term is 6 times the third term.

Options

A. 11,2
B. 2,11
C. 3,12
D. 12,3

Answer

To solve the problem, we need to find the values of \( n \) and \( x \) in the expansion of \( (1 + x)^n \) given that the fifth term is four times the fourth term and the fourth term is six times the third term.

### Step-by-Step Solution:

1. **Identify the Terms**:
The \( r \)-th term in the expansion of \( (1 + x)^n \) is given by:
\[ T_r = \binom{n}{r-1} x^{r-1} \]
Thus, we can write:
- Fifth term \( T_5 = \binom{n}{4} x^4 \)
- Fourth term \( T_4 = \binom{n}{3} x^3 \)
- Third term \( T_3 = \binom{n}{2} x^2 \)

2. **Set Up the First Condition**:
We are given that the fifth term is four times the fourth term:
\[ T_5 = 4 T_4 \]
Substituting the expressions for \( T_5 \) and \( T_4 \):
\[ \binom{n}{4} x^4 = 4 \cdot \binom{n}{3} x^3 \]

3. **Simplify the First Condition**:
Dividing both sides by \( x^3 \) (assuming \( x \neq 0 \)):
\[ \binom{n}{4} x = 4 \cdot \binom{n}{3} \]
Using the formula for binomial coefficients:
\[ \frac{n!}{4!(n-4)!} x = 4 \cdot \frac{n!}{3!(n-3)!} \]
This simplifies to:
\[ \frac{n!}{4!(n-4)!} x = \frac{4n!}{6(n-3)!} \]
Canceling \( n! \) from both sides:
\[ \frac{x}{4!} = \frac{4}{6} \cdot \frac{1}{(n-3)(n-4)} \]
Thus:
\[ x = \frac{4 \cdot 4!}{6} \cdot (n-3)(n-4) \]
Simplifying gives:
\[ x = \frac{96}{6} (n-3)(n-4) = 16(n-3)(n-4) \]

4. **Set Up the Second Condition**:
We are also given that the fourth term is six times the third term:
\[ T_4 = 6 T_3 \]
Substituting the expressions for \( T_4 \) and \( T_3 \):
\[ \binom{n}{3} x^3 = 6 \cdot \binom{n}{2} x^2 \]

5. **Simplify the Second Condition**:
Dividing both sides by \( x^2 \) (assuming \( x \neq 0 \)):
\[ \binom{n}{3} x = 6 \cdot \binom{n}{2} \]
Using the formula for binomial coefficients:
\[ \frac{n!}{3!(n-3)!} x = 6 \cdot \frac{n!}{2!(n-2)!} \]
Canceling \( n! \):
\[ \frac{x}{3!} = 6 \cdot \frac{1}{2(n-2)} \]
Thus:
\[ x = 6 \cdot 3! \cdot \frac{1}{2(n-2)} = 36 \cdot \frac{1}{2(n-2)} = \frac{18}{n-2} \]

6. **Equate the Two Expressions for \( x \)**:
Now we have two expressions for \( x \):
\[ 16(n-3)(n-4) = \frac{18}{n-2} \]

7. **Cross Multiply and Solve for \( n \)**:
Cross multiplying gives:
\[ 16(n-3)(n-4)(n-2) = 18 \]
Expanding and simplifying will lead to a polynomial equation in \( n \).

8. **Find the Roots**:
Solve the polynomial equation to find the possible values of \( n \).

9. **Substitute Back to Find \( x \)**:
Once \( n \) is found, substitute back into either expression for \( x \) to find its value.