Question

If the sum of the coefficients of $x^7$ and $x^{14}$ in the expansion of $\left( \frac{1}{x^3} - x^4 \right)^n, x \neq 0,$ is zero, then the value of $n$ is _________.

Answer 1

Correct Answer: 21

Step 1: Understanding the Question:
We need to find the general term of the binomial expansion and determine the index $n$ for which the sum of coefficients of two specific powers of $x$ is zero.
Step 2: Key Formula or Approach:
General term $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ for expansion $(a+b)^n$.
Step 3: Detailed Explanation:
General term of $\left( x^{-3} - x^4 \right)^n$:
\[ T_{r+1} = \binom{n}{r} (x^{-3})^{n-r} (-x^4)^r = \binom{n}{r} (-1)^r x^{-3n+3r+4r} = \binom{n}{r} (-1)^r x^{7r-3n} \]
For $x^7$: $7r - 3n = 7 \implies r_1 = \frac{3n+7}{7} = \frac{3n}{7} + 1$.
For $x^{14}$: $7r - 3n = 14 \implies r_2 = \frac{3n+14}{7} = \frac{3n}{7} + 2$.
Coefficients are $C_1 = \binom{n}{r_1} (-1)^{r_1}$ and $C_2 = \binom{n}{r_2} (-1)^{r_2}$.
Given $C_1 + C_2 = 0 \implies \binom{n}{r_1} (-1)^{r_1} + \binom{n}{r_1+1} (-1)^{r_1+1} = 0$.
\[ \binom{n}{r_1} (-1)^{r_1} - \binom{n}{r_1+1} (-1)^{r_1} = 0 \implies \binom{n}{r_1} = \binom{n}{r_1+1} \]
By binomial property, $\binom{n}{x} = \binom{n}{y} \implies x = y$ or $x+y = n$.
Since $r_1 \neq r_1+1$, we have $r_1 + (r_1+1) = n \implies 2r_1 + 1 = n$.
Substitute $r_1 = \frac{3n+7}{7}$:
\[ 2 \left( \frac{3n+7}{7} \right) + 1 = n \]
\[ \frac{6n + 14 + 7}{7} = n \]
\[ 6n + 21 = 7n \implies n = 21 \]
Step 4: Final Answer:
The value of $n$ is 21.