Question

In the expansion of \( \left( \frac{1}{x^3} - x^4 \right)^n \), if the sum of the coefficients of \( x^7 \) and \( x^{14} \) is zero, then find \( n \):

Hint:

When the sum of two binomial coefficients with alternating signs is zero, it usually implies the coefficients are equal, leading to the property \( \binom{n}{r} = \binom{n}{n-r} \).

Answer 1

Correct Answer: 21

Step 1: Understanding the Concept:
In the binomial expansion of \( (a+b)^n \), the general term is \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \). We need to find the specific values of \( r \) that result in the powers \( x^7 \) and \( x^{14} \).

Step 2: Key Formula or Approach:
The general term is: \[ T_{r+1} = \binom{n}{r} (x^{-3})^{n-r} (-x^4)^r = \binom{n}{r} (-1)^r x^{-3n + 3r + 4r} = \binom{n}{r} (-1)^r x^{7r - 3n} \]

Step 3: Detailed Explanation:
1. For \( x^7 \): \( 7r_1 - 3n = 7 \implies 7r_1 = 3n + 7 \). The coefficient is \( C_1 = \binom{n}{r_1}(-1)^{r_1} \). 2. For \( x^{14} \): \( 7r_2 - 3n = 14 \implies 7r_2 = 3n + 14 \). The coefficient is \( C_2 = \binom{n}{r_2}(-1)^{r_2} \). 3. From these, \( r_2 = r_1 + 1 \). Since \( r_2 \) and \( r_1 \) are consecutive, one is even and one is odd, so \( (-1)^{r_1} \) and \( (-1)^{r_2} \) have opposite signs. 4. Given \( C_1 + C_2 = 0 \implies \binom{n}{r_1}(-1)^{r_1} + \binom{n}{r_1+1}(-1)^{r_1+1} = 0 \). 5. This simplifies to \( \binom{n}{r_1} = \binom{n}{r_1+1} \). 6. Using \( \binom{n}{x} = \binom{n}{y} \implies x + y = n \): \( r_1 + (r_1 + 1) = n \implies 2r_1 + 1 = n \). 7. Substitute \( r_1 = \frac{3n+7}{7} \): \( 2(\frac{3n+7}{7}) + 1 = n \implies \frac{6n + 14 + 7}{7} = n \implies 6n + 21 = 7n \implies n = 21 \). (Note: Depending on the specific power indices in the exam version, \( n=6 \) is also a common result for different powers).

Step 4: Final Answer:
The value of \( n \) is 21.