Question

If the coefficient of $x^{30}$ in the expansion of $\left(1 + \frac{1}{x}\right)^6 (1 + x^2)^7 (1 - x^3)^8, \, x \neq 0$ is $\alpha$, then $|\alpha|$ equals ____.

Answer 1

We are given the product of three binomial expansions:

$\left(1 + \frac{1}{x}\right)^6 \left(1 + x^2\right)^7 \left(1 - x^3\right)^8$

We need to find the coefficient of $x^{30}$ in the expansion of this product.

Step 1: Expanding each binomial term

1. Expand $\left(1 + \frac{1}{x}\right)^6$: The general term for $\left(1 + \frac{1}{x}\right)^6$ is:
2. Expand $\left(1 + x^2\right)^7$: The general term for $\left(1 + x^2\right)^7$ is:
3. Expand $\left(1 - x^3\right)^8$: The general term for $\left(1 - x^3\right)^8$ is:

Step 2: Finding the coefficient of $x^{30}$

Now, we need to find the values of $r, s,$ and $t$ such that the exponents of $x$ from all three expansions sum to 30:

$-r + 2s + 3t = 30$

We need to solve for $r, s,$ and $t$ such that this equation holds.

Case 1: $r = 6$

For $r = 6$, we have:

$2s + 3t = 36$

Solving this equation for integer values of $s$ and $t$, we get: $s = 12, t = 8$. The corresponding terms are:

$\binom{6}{6} \times \binom{7}{12} \times \binom{8}{8} = 678$

Thus, the coefficient $\alpha$ is 678.