سوال ۱۱۱ کنکور ارشد علوم کامپیوتر ۱۴۰۲

We need the coefficient of $x^{9}$ in the power-series expansion of

$f(x) = \frac{1 + x^{3}}{(1 + x)^{3}}.$

Simplify first. Factor the numerator using $1 + x^{3} = (1 + x)(1 - x + x^{2})$:

$f(x) = \frac{(1 + x)(1 - x + x^{2})}{(1 + x)^{3}} = \frac{1 - x + x^{2}}{(1 + x)^{2}}.$

Expand $(1+x)^{-2}$ with the binomial series $\dfrac{1}{(1+x)^{2}} = \sum_{n \ge 0} \binom{n+1}{1}(-1)^{n} x^{n} = \sum_{n \ge 0} (-1)^{n}(n+1)\, x^{n}.$

Multiplying by $1 - x + x^{2}$, the coefficient of $x^{n}$ (for $n \ge 2$) is

$a_{n} = (-1)^{n}(n+1) - (-1)^{n-1}n + (-1)^{n-2}(n-1).$

Since $(-1)^{n-1} = (-1)^{n+1} = -(-1)^{n}$ and $(-1)^{n-2} = (-1)^{n}$, this collapses to

$a_{n} = (-1)^{n}\big[(n+1) + n + (n-1)\big] = (-1)^{n}\,(3n).$

Evaluate at $n = 9$:

$a_{9} = (-1)^{9}\cdot 3\cdot 9 = -27.$

A quick sanity check of the leading terms agrees: $f(x) = 1 - 3x + 6x^{2} - 9x^{3} + 12x^{4} - \cdots$, i.e. $a_{n} = (-1)^{n} 3n$, so $a_{9} = -27$.

The correct choice is option 2.

Pitfall: The decoy options $\pm(2^{9}+2)$ tempt you to treat $x^{9}$ as an exponential term. There is nothing exponential here — the answer grows only linearly in $n$, so $|a_{9}| = 27$, not $\sim 2^{9}$.