Question

If the coefficient of \(x^4\) in the binomial expansion of \((4x + a)^7\) is -1120, then the value of \(a\) is equal to

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{-1}{2}\)
• D. \(\frac{1}{8}\)
• E. \(\frac{-1}{8}\)

Hint:

Always ensure the exponent of the variable matches the requested power before solving for the unknown constant. Here, \(a^3 = -1/8\) directly implies \(a\) must be negative.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use the general term of the binomial expansion \((x + y)^n\) to find the specific term containing \(x^4\).

Step 2: Key Formula or Approach:
The general term of \((x+y)^n\) is \(T_{r+1} = ^nC_r x^{n-r} y^r\).

Step 3: Detailed Explanation:
For \((4x + a)^7\), the general term is:
\[ T_{r+1} = ^7C_r (4x)^{7-r} a^r = ^7C_r \cdot 4^{7-r} \cdot x^{7-r} \cdot a^r \]
We need the coefficient of \(x^4\), so we set the exponent of \(x\) to 4:
\[ 7 - r = 4 \implies r = 3 \]
Substituting \(r = 3\) into the coefficient part:
\[ \text{Coefficient} = ^7C_3 \cdot 4^{7-3} \cdot a^3 = ^7C_3 \cdot 4^4 \cdot a^3 \]
Given that the coefficient is -1120:
\[ ^7C_3 \cdot 256 \cdot a^3 = -1120 \]
Calculate \(^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\).
\[ 35 \cdot 256 \cdot a^3 = -1120 \]
\[ 8960 \cdot a^3 = -1120 \]
\[ a^3 = \frac{-1120}{8960} \]
\[ a^3 = -\frac{1}{8} \]
Taking the cube root:
\[ a = -\frac{1}{2} \]

Step 4: Final Answer:
The value of \(a\) is \(-\frac{1}{2}\).