Key Concept: Binomial Expansion and Middle Term

The binomial theorem provides a formula for expanding expressions of the form $(a+b)^n$. The general term, ${T_{r+1}}$, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^n{C_r} a^{n-r} b^r$ where ${}^n{C_r} = {n! \over r!(n-r)!}$.

To find the middle term(s) in a binomial expansion:

• If 'n' is even, there is only one middle term, which is the $({\frac{n}{2}} + 1)^{\text{th}}$ term.
• If 'n' is odd, there are two middle terms, which are the $({\frac{n+1}{2}})^{\text{th}}$ term and the $({\frac{n+3}{2}})^{\text{th}}$ term.

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Step 1: Find the coefficient of the middle term in the expansion of ${(1 + \alpha x)^4}$

1. Identify 'n' and terms 'a' and 'b': For the expression ${(1 + \alpha x)^4}$, we have $n=4$. Here, $a=1$ and $b=\alpha x$.

2. Determine the position of the middle term: Since $n=4$ is an even number, there is only one middle term. Its position is given by $({\frac{n}{2}} + 1)^{\text{th}}$ term. Middle term position = $({\frac{4}{2}} + 1)^{\text{th}} = (2 + 1)^{\text{th}} = 3^{\text{rd}}$ term. Therefore, we need to find ${T_3}$.

3. Apply the general term formula: For ${T_3}$, we set $r+1=3$, which means $r=2$. Substituting $n=4$, $a=1$, $b=\alpha x$, and $r=2$ into the general term formula: $T_{2+1} = {}^4{C_2} (1)^{4-2} (\alpha x)^2$ $T_3 = {}^4{C_2} (1)^2 (\alpha^2 x^2)$ $T_3 = {}^4{C_2} \alpha^2 x^2$

4. Calculate the binomial coefficient ${}^4{C_2}$: ${}^4{C_2} = {4! \over 2!(4-2)!} = {4! \over 2!2!} = {4 \times 3 \times 2 \times 1 \over (2 \times 1)(2 \times 1)} = {24 \over 4} = 6$

5. Identify the coefficient of the middle term: From ${T_3} = 6 \alpha^2 x^2$, the coefficient of the middle term in ${(1 + \alpha x)^4}$ is $6 \alpha^2$.

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Step 2: Find the coefficient of the middle term in the expansion of ${(1 - \alpha x)^6}$

1. Identify 'n' and terms 'a' and 'b': For the expression ${(1 - \alpha x)^6}$, we have $n=6$. Here, $a=1$ and $b=-\alpha x$. Self-correction/Tip: Pay close attention to the sign of the second term 'b'. If it's negative, include the negative sign when substituting into the formula.

2. Determine the position of the middle term: Since $n=6$ is an even number, there is only one middle term. Its position is given by $({\frac{n}{2}} + 1)^{\text{th}}$ term. Middle term position = $({\frac{6}{2}} + 1)^{\text{th}} = (3 + 1)^{\text{th}} = 4^{\text{th}}$ term. Therefore, we need to find ${T_4}$.

3. Apply the general term formula: For ${T_4}$, we set $r+1=4$, which means $r=3$. Substituting $n=6$, $a=1$, $b=-\alpha x$, and $r=3$ into the general term formula: $T_{3+1} = {}^6{C_3} (1)^{6-3} (-\alpha x)^3$ $T_4 = {}^6{C_3} (1)^3 (-\alpha^3 x^3)$ $T_4 = {}^6{C_3} (-\alpha^3 x^3)$

4. Calculate the binomial coefficient ${}^6{C_3}$: ${}^6{C_3} = {6! \over 3!(6-3)!} = {6! \over 3!3!} = {6 \times 5 \times 4 \times 3 \times 2 \times 1 \over (3 \times 2 \times 1)(3 \times 2 \times 1)} = {720 \over 36} = 20$

5. Identify the coefficient of the middle term: From ${T_4} = 20 (-\alpha^3 x^3) = -20 \alpha^3 x^3$, the coefficient of the middle term in ${(1 - \alpha x)^6}$ is $-20 \alpha^3$.

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Step 3: Equate the coefficients and solve for $\alpha$

The problem states that the coefficients of the middle terms in both expansions are the same. From Step 1, the coefficient for ${(1 + \alpha x)^4}$ is $6 \alpha^2$. From Step 2, the coefficient for ${(1 - \alpha x)^6}$ is $-20 \alpha^3$.

Setting them equal: $6 \alpha^2 = -20 \alpha^3$

To solve for $\alpha$, we can rearrange the equation: $20 \alpha^3 + 6 \alpha^2 = 0$ Factor out the common term, which is $2 \alpha^2$: $2 \alpha^2 (10 \alpha + 3) = 0$

This equation gives two possibilities for $\alpha$:

1. $2 \alpha^2 = 0 \Rightarrow \alpha^2 = 0 \Rightarrow \alpha = 0$
2. $10 \alpha + 3 = 0 \Rightarrow 10 \alpha = -3 \Rightarrow \alpha = -{3 \over 10}$

If $\alpha = 0$, then both coefficients would be zero ($6(0)^2 = 0$ and $-20(0)^3 = 0$), making them equal. However, looking at the given options, we are looking for a non-zero value of $\alpha$. Therefore, we choose the second solution.

$\alpha = -{3 \over 10}$

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Tips and Common Mistakes

• Sign Errors: A very common mistake is to forget the negative sign when the second term in the binomial is negative, like in $(1 - \alpha x)^6$. Always treat the 'b' term as its entire value, including its sign (e.g., $b = -\alpha x$). When raised to an odd power, a negative base remains negative.
• Counting Terms: Carefully determine the position of the middle term(s). For an even 'n', there's one middle term at $(n/2 + 1)$. For an odd 'n', there are two middle terms at $(n+1)/2$ and $(n+3)/2$.
• Simplifying ${}^n{C_r}$: Ensure correct calculation of binomial coefficients.
• Solving Algebraic Equations: When solving for $\alpha$, avoid simply dividing by $\alpha^2$ without first considering the case where $\alpha = 0$. Factoring out the common term ($2\alpha^2$ in this case) is a safer approach to ensure all possible solutions are found. In multiple-choice questions, often only the non-zero solutions are provided.

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Summary

By correctly identifying the middle terms for each binomial expansion and calculating their coefficients while carefully handling signs, we established an equation relating the coefficients. Solving this equation led us to the value of $\alpha = -{3 \over 10}$. This problem emphasizes the importance of accurately applying the binomial theorem's general term formula and paying close attention to algebraic details, especially signs.