Find Power n for Zero Sum of Coefficients

To find the value of $n$, we start by determining the general term in the binomial expansion of $\left(\frac{1}{x^3} - x^4\right)^n$.

Using the binomial theorem, the general term $T_{r+1}$ is given by: $T_{r+1} = \binom{n}{r} \left(\frac{1}{x^3}\right)^{n-r} \left(-x^4\right)^r$

Simplifying the powers of $x$: $T_{r+1} = \binom{n}{r} \left(x^{-3}\right)^{n-r} (-1)^r \left(x^4\right)^r$ $T_{r+1} = (-1)^r \binom{n}{r} x^{-3(n-r) + 4r}$ $T_{r+1} = (-1)^r \binom{n}{r} x^{7r - 3n}$

Step 1: Find the coefficient of $x^7$

To find the coefficient of $x^7$, we set the exponent of $x$ equal to $7$: $7r_1 - 3n = 7 \implies r_1 = \frac{3n + 7}{7}$

The coefficient of $x^7$ is: $C_1 = (-1)^{r_1} \binom{n}{r_1}$

Step 2: Find the coefficient of $x^{14}$

To find the coefficient of $x^{14}$, we set the exponent of $x$ equal to $14$: $7r_2 - 3n = 14 \implies r_2 = \frac{3n + 14}{7} = r_1 + 1$

The coefficient of $x^{14}$ is: $C_2 = (-1)^{r_2} \binom{n}{r_2} = (-1)^{r_1 + 1} \binom{n}{r_1 + 1}$

Step 3: Use the given condition

We are given that the sum of the coefficients of $x^7$ and $x^{14}$ is zero: $C_1 + C_2 = 0$ $(-1)^{r_1} \binom{n}{r_1} + (-1)^{r_1 + 1} \binom{n}{r_1 + 1} = 0$ $(-1)^{r_1} \left[ \binom{n}{r_1} - \binom{n}{r_1 + 1} \right] = 0$

Since $(-1)^{r_1} \neq 0$, we have: $\binom{n}{r_1} = \binom{n}{r_1 + 1}$

Using the identity $\binom{n}{a} = \binom{n}{b} \implies a + b = n$ (for $a \neq b$): $r_1 + (r_1 + 1) = n$ $2r_1 + 1 = n$

Step 4: Solve for $n$

Substitute $r_1 = \frac{3n + 7}{7}$ into the equation: $2\left(\frac{3n + 7}{7}\right) + 1 = n$ $\frac{6n + 14}{7} + 1 = n$ $6n + 14 + 7 = 7n$ $6n + 21 = 7n$ $n = 21$

Thus, the value of $n$ is 21.