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 are given a binomial expansion and a condition that the sum of the coefficients for two specific powers of $x$ is zero. We need to use this condition to find the value of the exponent $n$.
Step 2: Key Formula or Approach:
The general term (the $(r+1)$-th term) in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. We will apply this formula, solve for the powers of $x$, and then use the given condition on the coefficients.
Step 3: Detailed Explanation:
The given expression is $\left( x^{-3} - x^4 \right)^n$. The general term is:
\[ T_{r+1} = \binom{n}{r} (x^{-3})^{n-r} (-x^4)^r = \binom{n}{r} (-1)^r x^{-3n+3r} x^{4r} = \binom{n}{r} (-1)^r x^{7r-3n} \] The coefficient of the term with $x^{7r-3n}$ is $\binom{n}{r} (-1)^r$.
For the coefficient of $x^7$:
We set the exponent equal to 7: $7r - 3n = 7 \implies 7r = 3n+7 \implies r_1 = \frac{3n+7}{7}$.
The coefficient is $C_1 = \binom{n}{r_1} (-1)^{r_1}$.
For the coefficient of $x^{14$:}
We set the exponent equal to 14: $7r - 3n = 14 \implies 7r = 3n+14 \implies r_2 = \frac{3n+14}{7}$.
The coefficient is $C_2 = \binom{n}{r_2} (-1)^{r_2}$.
We can see that $r_2 = \frac{3n+7+7}{7} = r_1 + 1$.
The given condition is $C_1 + C_2 = 0$.
\[ \binom{n}{r_1} (-1)^{r_1} + \binom{n}{r_1+1} (-1)^{r_1+1} = 0 \] Factor out $(-1)^{r_1}$:
\[ (-1)^{r_1} \left[ \binom{n}{r_1} - \binom{n}{r_1+1} \right] = 0 \] This implies $\binom{n}{r_1} = \binom{n}{r_1+1}$.
Using the property that if $\binom{n}{k} = \binom{n}{j}$ and $k \neq j$, then $k+j=n$.
Here, we have $r_1 + (r_1+1) = n \implies 2r_1 + 1 = n$.
Now substitute the expression for $r_1$ in terms of $n$:
\[ 2 \left( \frac{3n+7}{7} \right) + 1 = n \] \[ \frac{6n+14}{7} + 1 = n \] \[ 6n + 14 + 7 = 7n \] \[ 6n + 21 = 7n \implies n = 21 \] Step 4: Final Answer:
The value of $n$ is 21.