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:
The general term in $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. We first find $n$ using the given coefficient.

Step 2: Detailed Explanation:
1. Find $n$:
In $(2+x)^n$, $T_{r+1} = \binom{n}{r} 2^{n-r} x^r$.
For $x^3$, $r=3$. Coeff $= \binom{n}{3} 2^{n-3} = 160$.
Test small values of $n$:
If $n=6$: $\binom{6}{3} 2^{6-3} = 20 \times 2^3 = 20 \times 8 = 160$. Correct. So, $n=6$.
2. Find coefficient of $x^6$ in $(2 - x^2)^6$:
General term: $T_{r+1} = \binom{6}{r} 2^{6-r} (-x^2)^r = \binom{6}{r} 2^{6-r} (-1)^r x^{2r}$.
We want $x^6$, so $2r = 6 \implies r = 3$.
Coefficient $= \binom{6}{3} 2^{6-3} (-1)^3$
Coefficient $= 20 \times 2^3 \times (-1) = 20 \times 8 \times (-1) = -160$.

Step 3: Final Answer:
The coefficient is -160.