The coefficient of the middle term in the binomial expansion in powers of `x` of`(1+alphax)^4` and of`(1-alphax)^6` is the same, if `alpha` equals `-5/3` b. `(10)/3` c. `-3/(10)` d. `3/5`

Options

A. `-(5)/(3)`
B. `(3)/(5) `
C. `-(3)/(10)`
D. `(10)/(3)`

Answer

To find the value of \(\alpha\) such that the coefficient of the middle term in the binomial expansions of \((1 + \alpha x)^4\) and \((1 - \alpha x)^6\) are the same, we will follow these steps:

### Step 1: Identify the middle term in the binomial expansion

For the binomial expansion of \((1 + \alpha x)^n\), the middle term can be found using the formula:
\[ T_k = \binom{n}{k} (1)^{n-k} (\alpha x)^k \]
where \(k = \frac{n}{2}\) if \(n\) is even, and \(k = \frac{n-1}{2}\) and \(k = \frac{n+1}{2}\) if \(n\) is odd.

**Hint:** Determine whether \(n\) is even or odd to find the correct middle term.

### Step 2: Calculate the middle term for \((1 + \alpha x)^4\)

Here, \(n = 4\) (which is even), so the middle term is:
\[ T_2 = \binom{4}{2} (1)^{4-2} (\alpha x)^2 = \binom{4}{2} \alpha^2 x^2 \]
Calculating \(\binom{4}{2}\):
\[ \binom{4}{2} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6 \]
Thus, the middle term becomes:
\[ T_2 = 6 \alpha^2 x^2 \]

**Hint:** Remember to calculate the binomial coefficient accurately.

### Step 3: Calculate the middle term for \((1 - \alpha x)^6\)

Now, for \(n = 6\) (which is also even), the middle term is:
\[ T_3 = \binom{6}{3} (1)^{6-3} (-\alpha x)^3 = \binom{6}{3} (-\alpha)^3 x^3 \]
Calculating \(\binom{6}{3}\):
\[ \binom{6}{3} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Thus, the middle term becomes:
\[ T_3 = 20 (-\alpha)^3 x^3 = -20 \alpha^3 x^3 \]

**Hint:** Pay attention to the sign when dealing with negative coefficients.

### Step 4: Set the coefficients of the middle terms equal

Since we want the coefficients of the middle terms to be the same, we set:
\[ 6 \alpha^2 = -20 \alpha^3 \]

**Hint:** Rearranging the equation can help simplify the problem.

### Step 5: Rearranging the equation

Rearranging gives:
\[ 20 \alpha^3 + 6 \alpha^2 = 0 \]
Factoring out \(\alpha^2\):
\[ \alpha^2 (20 \alpha + 6) = 0 \]

**Hint:** Factor out common terms to simplify the equation.

### Step 6: Solve for \(\alpha\)

This gives us two cases:
1. \(\alpha^2 = 0 \Rightarrow \alpha = 0\)
2. \(20 \alpha + 6 = 0 \Rightarrow 20 \alpha = -6 \Rightarrow \alpha = -\frac{6}{20} = -\frac{3}{10}\)

**Hint:** Check each case to ensure you find all possible solutions.

### Final Answer

The value of \(\alpha\) is:
\[ \alpha = -\frac{3}{10} \]