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

Correct Answer: 678

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.