Question

If the coefficient of $x^3$ in the binomial expansion of $(2 + x)^n$ is 160, then the coefficient of $x^6$ in the binomial expansion of $(2 - x^2)^n$ is

• A. 160
• B. 320
• C. -160
• D. -320
• E. -960

Hint:

When coefficient values like 160 appear with powers of 2, powers of 10 or factors of 20 are common. Testing $n=5$ or $n=6$ is often faster than solving the cubic equation $\frac{n(n-1)(n-2)}{6} 2^{n-3} = 160$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
First, use the given coefficient to determine the value of \(n\).
Then, evaluate the target coefficient in the second expansion using that \(n\).
Step 2: Key Formula or Approach:
The general term in the expansion of \((a + b)^n\) is \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).
For \((2 + x)^n\), the term with \(x^3\) corresponds to \(r=3\).
Step 3: Detailed Explanation:
Find \(n\):
The term with \(x^3\) is \(\binom{n}{3} 2^{n-3} x^3\).
We are given \(\binom{n}{3} 2^{n-3} = 160\).
By testing small integer values for \(n\):
If \(n = 6\): \(\binom{6}{3} 2^{6-3} = 20 \times 2^3 = 20 \times 8 = 160\).
So, \(n = 6\).
Now find the coefficient of \(x^6\) in \((2 - x^2)^6\).
The general term is \(\binom{6}{r} 2^{6-r} (-x^2)^r\).
We want the power of \(x\) to be 6, so \(2r = 6 \implies r = 3\).
Substitute \(r=3\):
Coefficient = \(\binom{6}{3} 2^{6-3} (-1)^3\)
\[ = 20 \times 8 \times (-1) = -160 \] Step 4: Final Answer:
The coefficient is -160.