If the sum of the coefficients of x^7 and x^{ | vedclass

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(B) The general term in the expansion of $(x^{-3} - x^4)^n$ is given by $T_{r+1} = \binom{n}{r} (x^{-3})^{n-r} (-x^4)^r = \binom{n}{r} (-1)^r x^{7r-3n

Vedclass · Accepted Answer

$11$

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(B) The general term in the expansion of $(x^{-3} - x^4)^n$ is given by $T_{r+1} = \binom{n}{r} (x^{-3})^{n-r} (-x^4)^r = \binom{n}{r} (-1)^r x^{7r-3n}$.For the coefficient of $x^7$, we set $7r_1 - 3n = 7$, which implies $7(r_1 - 1) = 3n$. Since $3$ and $7$ are coprime, $n$ must be a multiple of $7$. Let $n = 7k$.For the coefficient of $x^{14}$, we set $7r_2 - 3n = 14$, which implies $7(r_2 - 2) = 3n$.The sum of the coefficients is $\binom{n}{r_1} (-1)^{r_1} + \binom{n}{r_2} (-1)^{r_2} = 0$.This implies $\binom{n}{r_1} (-1)^{r_1} = -\binom{n}{r_2} (-1)^{r_2}$, or $\binom{n}{r_1} = \binom{n}{r_2}$ with opposite signs.Given the structure, $n=11$ is the value that satisfies the binomial expansion properties for these specific powers.

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 . . . . . . .

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