Question

If the coefficient of $y^{3}$ in the binomial expansion of $\left(2\alpha-\frac{y}{2}\right)^{8}$ is -7, then the value of $\alpha$ is equal to

• A. $\frac{1}{2}$
• B. 2
• C. 3
• D. 6
• E. 8

Hint:

Logic Tip: Always clearly separate constants from variables when applying the general term formula. Pay special attention to negative signs inside the binomial, as they are raised to the power $r$ and frequently cause calculation errors.

Answer 1

The Correct Option is A

Concept:
The general term $T_{r+1}$ in the binomial expansion of $(x + y)^n$ is given by: $$T_{r+1} = \binom{n}{r} x^{n-r} y^r$$ We can use this to find the coefficient of any specific power of the variable.
Step 1: Write the general term for the given expansion.
For the expression $\left(2\alpha - \frac{y}{2}\right)^8$, the general term is: $$T_{r+1} = \binom{8}{r} (2\alpha)^{8-r} \left(-\frac{y}{2}\right)^r$$
Step 2: Isolate the variable part to find the required r.
We are looking for the coefficient of $y^3$. Let's separate the constants from the variables in the general term: $$T_{r+1} = \binom{8}{r} (2\alpha)^{8-r} \left(-\frac{1}{2}\right)^r y^r$$ To find the term containing $y^3$, we set the exponent of $y$ equal to 3: $$r = 3$$
Step 3: Calculate the coefficient for r = 3.
Substitute $r = 3$ back into the constant part of the general term: $$\text{Coefficient} = \binom{8}{3} (2\alpha)^{8-3} \left(-\frac{1}{2}\right)^3$$ $$\text{Coefficient} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} \cdot (2\alpha)^5 \cdot \left(-\frac{1}{8}\right)$$ $$\text{Coefficient} = 56 \cdot 32\alpha^5 \cdot \left(-\frac{1}{8}\right)$$ $$\text{Coefficient} = 56 \cdot (-4) \cdot \alpha^5 = -224\alpha^5$$
Step 4: Equate to the given value and solve for $\alpha$.
We are given that this coefficient equals -7: $$-224\alpha^5 = -7$$ $$\alpha^5 = \frac{-7}{-224}$$ $$\alpha^5 = \frac{1}{32}$$ Taking the fifth root of both sides gives: $$\alpha = \frac{1}{2}$$