Question

In the binomial expansion of \( (2x + \alpha)^8 \), the co-efficients of \( x^2 \) and \( x^3 \) are equal. Then the value of \( \alpha \) is equal to

• A. \(2 \)
• B. \( \frac{1}{4} \)
• C. \(4 \)
• D. \( \frac{1}{2} \)
• E. \(3 \)

Hint:

Compare coefficients carefully by matching powers of \(x\) using binomial term formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem requires finding specific terms in a binomial expansion and equating their coefficients.
Step 2: Key Formula or Approach:
The general term (the (r+1)\(^{th}\) term) in the expansion of \((a+b)^n\) is given by: \[ T_{r+1} = {}^nC_r a^{n-r} b^r \] For our problem, \(a = 2x\), \(b = \alpha\), and \(n = 8\). So, the general term is \(T_{r+1} = {}^8C_r (2x)^{8-r} (\alpha)^r = {}^8C_r 2^{8-r} \alpha^r x^{8-r}\).
Step 3: Detailed Explanation:
1. Find the coefficient of x\(^3\):
To get the term with x\(^3\), we need the power of x to be 3. So, \(8-r = 3\), which implies \(r = 5\). The term is \(T_{5+1} = T_6\). The coefficient is \({}^8C_5 \cdot 2^{8-5} \cdot \alpha^5\). \[ \text{Coeff(x}^3\text{)} = {}^8C_3 \cdot 2^3 \cdot \alpha^5 \quad (\text{since } {}^8C_5 = {}^8C_3) \] \[ = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} \cdot 8 \cdot \alpha^5 = 56 \cdot 8 \cdot \alpha^5 = 448 \alpha^5 \] 2. Find the coefficient of x\(^2\):
To get the term with x\(^2\), we need \(8-r = 2\), which implies \(r = 6\). The term is \(T_{6+1} = T_7\). The coefficient is \({}^8C_6 \cdot 2^{8-6} \cdot \alpha^6\). \[ \text{Coeff(x}^2\text{)} = {}^8C_2 \cdot 2^2 \cdot \alpha^6 \quad (\text{since } {}^8C_6 = {}^8C_2) \] \[ = \frac{8 \cdot 7}{2 \cdot 1} \cdot 4 \cdot \alpha^6 = 28 \cdot 4 \cdot \alpha^6 = 112 \alpha^6 \] 3. Equate the coefficients:
We are given that the coefficients are equal: \[ 448 \alpha^5 = 112 \alpha^6 \] Assuming \(\alpha \neq 0\), we can divide both sides by \(112\alpha^5\): \[ \frac{448}{112} = \frac{112 \alpha^6}{112 \alpha^5} \] \[ 4 = \alpha \] Step 4: Final Answer:
The value of \(\alpha\) is 4.