Key Concepts and Formulas

This problem primarily involves two key mathematical concepts:

1. Binomial Theorem: The general term $(T_{r+1})$ in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^n C_r a^{n-r} b^r$. The coefficient of this term is ${}^n C_r a^{n-r} b^r$, but when we refer to the "coefficient of the middle term," it usually implies the numerical part multiplying the variable, or the entire term if the variable is part of $b$.
   • For an expansion $(a+b)^n$, if $n$ is an even integer, there is exactly one middle term, which is the $(\frac{n}{2} + 1)^{\text{th}}$ term.
   • If $n$ is an odd integer, there are two middle terms, which are the $(\frac{n+1}{2})^{\text{th}}$ and $(\frac{n+3}{2})^{\text{th}}$ terms.
2. Arithmetic Progression (A.P.): A sequence of numbers is in A.P. if the difference between consecutive terms is constant. If three terms, say $A, B, C$, are in A.P., then the middle term is the arithmetic mean of the other two, i.e., $2B = A+C$. The common difference ($d$) is given by $d = B-A = C-B$.

Step-by-Step Derivation

1. Finding the Coefficients of the Middle Terms for Each Expansion

• Expansion 1: $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4}$ Here, the power is $n=4$ (an even number).

  • Explanation: Since $n$ is even, there is one middle term. Its position is $\frac{n}{2} + 1 = \frac{4}{2} + 1 = 3$. So, we need to find the coefficient of the $3^{\text{rd}}$ term ($T_3$).
  • For $T_3$, we use $r=2$ in the general term formula.
  • The term is $T_3 = {}^4 C_2 \left(\frac{1}{\sqrt{6}}\right)^{4-2} (\beta x)^{2}$.
  • The coefficient of the term (ignoring $x$) is ${}^4 C_2 \left(\frac{1}{\sqrt{6}}\right)^{2} (\beta)^{2}$.
  • Calculating ${}^4 C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
  • So, the coefficient is $6 \times \left(\frac{1}{6}\right) \times \beta^2 = \beta^2$.
  • First term of the A.P. ($A_1$): $\beta^2$

• Expansion 2: $(1-3 \beta x)^{2}$ Here, the power is $n=2$ (an even number).

  • Explanation: Similar to the first expansion, the middle term position is $\frac{n}{2} + 1 = \frac{2}{2} + 1 = 2$. We need the coefficient of the $2^{\text{nd}}$ term ($T_2$).
  • For $T_2$, we use $r=1$ in the general term formula.
  • The term is $T_2 = {}^2 C_1 (1)^{2-1} (-3 \beta x)^{1}$.
  • The coefficient of the term (ignoring $x$) is ${}^2 C_1 (1)^{1} (-3 \beta)^{1}$.
  • Calculating ${}^2 C_1 = 2$.
  • So, the coefficient is $2 \times 1 \times (-3\beta) = -6\beta$.
  • Second term of the A.P. ($A_2$): $-6\beta$

• Expansion 3: $\left(1-\frac{\beta}{2} x\right)^{6}$ Here, the power is $n=6$ (an even number).

  • Explanation: The middle term position is $\frac{n}{2} + 1 = \frac{6}{2} + 1 = 4$. We need the coefficient of the $4^{\text{th}}$ term ($T_4$).
  • For $T_4$, we use $r=3$ in the general term formula.
  • The term is $T_4 = {}^6 C_3 (1)^{6-3} \left(-\frac{\beta}{2} x\right)^{3}$.
  • The coefficient of the term (ignoring $x$) is ${}^6 C_3 (1)^{3} \left(-\frac{\beta}{2}\right)^{3}$.
  • Calculating ${}^6 C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
  • So, the coefficient is $20 \times 1 \times \left(-\frac{\beta^3}{8}\right) = -\frac{20 \beta^3}{8} = -\frac{5 \beta^3}{2}$.
  • Third term of the A.P. ($A_3$): $-\frac{5 \beta^3}{2}$

2. Forming the A.P. and Solving for $\beta$

• Explanation: The three coefficients $\beta^2$, $-6\beta$, and $-\frac{5 \beta^3}{2}$ form an A.P. We use the property $2B = A+C$.
• Substituting the terms: $2(-6\beta) = \beta^2 + \left(-\frac{5 \beta^3}{2}\right)$ $-12\beta = \beta^2 - \frac{5 \beta^3}{2}$
• Explanation: We need to solve this equation for $\beta$. Since we are given $\beta > 0$, we know $\beta \neq 0$, so we can safely divide the entire equation by $\beta$. This simplifies the cubic equation to a quadratic one. $-12 = \beta - \frac{5 \beta^2}{2}$
• Explanation: To eliminate the fraction and make the equation easier to solve, multiply every term by 2. $-24 = 2\beta - 5\beta^2$
• Explanation: Rearrange the terms to form a standard quadratic equation $ax^2+bx+c=0$. $5\beta^2 - 2\beta - 24 = 0$
• Explanation: Solve the quadratic equation for $\beta$. We can use factorization (splitting the middle term) or the quadratic formula. For factorization, we look for two numbers that multiply to $5 \times (-24) = -120$ and add to $-2$. These numbers are $-12$ and $10$. $5\beta^2 - 12\beta + 10\beta - 24 = 0$
• Explanation: Factor by grouping. $\beta(5\beta - 12) + 2(5\beta - 12) = 0$ $(5\beta + 2)(5\beta - 12) = 0$
• Explanation: This gives two possible values for $\beta$. $5\beta + 2 = 0 \Rightarrow \beta = -\frac{2}{5}$ $5\beta - 12 = 0 \Rightarrow \beta = \frac{12}{5}$
• Explanation: The problem states that $\beta > 0$. Therefore, we choose the positive value. $\beta = \frac{12}{5}$

3. Calculating the Common Difference ($d$) and the Final Expression

• Explanation: The common difference $d$ is defined as the difference between any two consecutive terms in the A.P. Let's use $d = A_2 - A_1$. $A_1 = \beta^2$ $A_2 = -6\beta$ So, $d = -6\beta - \beta^2 = -(6\beta + \beta^2)$
• Explanation: Now we need to evaluate the expression $50-\frac{2 d}{\beta^{2}}$. Substitute the expression for $d$ into this. $50 - \frac{2(-(6\beta + \beta^2))}{\beta^2}$
• Explanation: Simplify the expression. The two negative signs cancel out, and we can split the fraction. $50 + \frac{2(6\beta + \beta^2)}{\beta^2}$ $= 50 + \frac{12\beta + 2\beta^2}{\beta^2}$ $= 50 + \left(\frac{12\beta}{\beta^2} + \frac{2\beta^2}{\beta^2}\right)$ $= 50 + \frac{12}{\beta} + 2$ $= 52 + \frac{12}{\beta}$
• Explanation: Substitute the value of $\beta = \frac{12}{5}$ into this simplified expression. $52 + \frac{12}{\left(\frac{12}{5}\right)}$ $= 52 + 12 \times \frac{5}{12}$ $= 52 + 5$ $= 57$

Tips and Common Mistakes

• Careful with signs: Especially when dealing with negative terms in binomial expansions like $(1-3\beta x)^2$, ensure the negative sign is carried over with the term $-3\beta x$. For example, $(-3\beta x)^1 = -3\beta x$ and $(-x)^3 = -x^3$.
• Distinguish between term and coefficient: The problem asks for "coefficients of the middle terms," so make sure to extract only the numerical part (and parts involving $\beta$) from the binomial term, not including the variable $x$.
• Verify A.P. condition: Always use $2B = A+C$ for three terms in A.P. This is often less prone to error than calculating $d$ first and then verifying.
• Quadratic Equation Solving: Double-check your factorization or quadratic formula calculations to avoid algebraic errors.
• Constraints on variables: Pay attention to conditions like $\beta > 0$. This is crucial for selecting the correct solution if multiple values arise.
• Simplify before substituting: In the final calculation, simplifying the expression involving $d$ and $\beta^2$ before substituting the value of $\beta$ can often lead to much simpler arithmetic, as shown above ($52 + \frac{12}{\beta}$ vs. substituting into the full fraction).

Summary

We successfully found the coefficients of the middle terms for each binomial expansion by applying the binomial theorem. These coefficients were then used to form an arithmetic progression, allowing us to establish an equation for $\beta$. Solving this quadratic equation and adhering to the condition $\beta > 0$, we found $\beta = \frac{12}{5}$. Finally, by expressing the common difference $d$ in terms of $\beta$ and substituting it into the given expression $50-\frac{2 d}{\beta^{2}}$, we simplified the expression and calculated its value to be $57$. The key takeaway is the importance of systematically applying binomial theorem properties and arithmetic progression definitions, coupled with careful algebraic manipulation.