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

Concept: General term: \[ T_{k+1} = {^nC_k}(a)^{n-k}(b)^k \]

Step 1: Coefficient of \(x^2\).
\[ x^2 \Rightarrow k=2 \] \[ {^8C_2}(2x)^2 \alpha^{6} = {^8C_2} \cdot 2^2 \cdot \alpha^6 \cdot x^2 \] Coefficient: \[ {^8C_2} \cdot 4 \cdot \alpha^6 \]

Step 2: Coefficient of \(x^3\).
\[ k=3 \] \[ {^8C_3}(2x)^3 \alpha^{5} \] Coefficient: \[ {^8C_3} \cdot 8 \cdot \alpha^5 \]

Step 3: Equate coefficients.
\[ {^8C_2} \cdot 4 \cdot \alpha^6 = {^8C_3} \cdot 8 \cdot \alpha^5 \]

Step 4: Simplify.
\[ 28 \cdot 4 \cdot \alpha^6 = 56 \cdot 8 \cdot \alpha^5 \] \[ 112 \alpha^6 = 448 \alpha^5 \] \[ \alpha = 4 \]