<p>Find the coefficient of <em>i</em><sup>8</sup> in the expansion of <span>\((1 + 2r^2 - r^3)^9\)</span>.</p>
Step-by-Step Solution
Key Concept: We need to find all ways to select terms from the trinomial expansion that result in $r^8$ as the power of $r$. Use the multinomial theorem to identify which combinations of $(1)$, $(2r^2)$, and $(-r^3)$ yield $r^8$.
<p><strong>Step 1: Set up the multinomial expansion</strong></p><p>Using the multinomial theorem, $(1 + 2r^2 - r^3)^9$ expands as:</p><p>$$\sum \frac{9!}{a!b!c!}(1)^a(2r^2)^b(-r^3)^c$$</p><p>where $a + b + c = 9$ and the coefficient is $\frac{9!}{a!b!c!} \cdot 2^b \cdot (-1)^c$.</p><p><strong>Step 2: Find the power of $r$</strong></p><p>The power of $r$ in each term is $2b + 3c$. We need:</p><p>$$2b + 3c = 8$$</p><p><strong>Step 3: Find all valid combinations</strong></p><p>With constraints $a + b + c = 9$ and $2b + 3c = 8$:</p><p>From $2b + 3c = 8$, we get $b = \frac{8-3c}{2}$ (which must be a non-negative integer).</p><p>For $c = 0$: $b = 4$, so $a = 9 - 4 - 0 = 5$ ✓</p><p>For $c = 1$: $b = \frac{8-3}{2} = 2.5$ (not an integer) ✗</p><p>For $c = 2$: $b = \frac{8-6}{2} = 1$, so $a = 9 - 1 - 2 = 6$ ✓</p><p>For $c \geq 3$: $2b + 3c \geq 9 > 8$ (no solutions)</p><p><strong>Step 4: Calculate the coefficient for $(a,b,c) = (5,4,0)$</strong></p><p>$$\frac{9!}{5!4!0!} \cdot 2^4 \cdot (-1)^0 = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} \cdot 16 = 126 \cdot 16 = 2016$$</p><p><strong>Step 5: Calculate the coefficient for $(a,b,c) = (6,1,2)$</strong></p><p>$$\frac{9!}{6!1!2!} \cdot 2^1 \cdot (-1)^2 = \frac{9 \cdot 8 \cdot 7}{2 \cdot 1} \cdot 2 \cdot 1 = 252 \cdot 2 = 504$$</p><p><strong>Step 6: Add the contributions</strong></p><p>Total coefficient of $r^8$ is:</p><p>$$2016 + 504 = 2520$$</p><p><strong>∴ Answer: 2520</strong></p>
Correct Answer: 2520