Question:

Step 1: Identify the coefficient to be found. The given expansion is $$(1 + 2x + 3x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \cdots + a_{20} x^{20}$$ We are tasked with finding the coefficient $a_1$. Step 2: Apply the multinomial theorem. To obtain a term with $x^1$ in the expansion of $(1 + 2x + 3x^2)^{10}$, we must select the term $2x$ exactly once from one of the ten factors $(1 + 2x + 3x^2)$ and the term $1$ from the remaining nine factors. The term $3x^2$ cannot be selected, as it would result in $x^2$ or higher powers. The number of ways to choose the factor contributing $2x$ is $\binom{10}{1}$. Each such selection contributes $ (2x)^1 \cdot (1)^9 = 2x $. Therefore, the coefficient of $x$ is: $$a_1 = \binom{10}{1} \cdot 2 = 10 \cdot 2 = 20$$ Step 3: Alternatively, use differentiation. The coefficient $a_1$ can be found by evaluating the first derivative of the function at $x=0$. Let $f(x) = (1 + 2x + 3x^2)^{10}$. Then $a_1 = \frac{f'(0)}{1!}$. First, compute the derivative $f'(x)$: $$f'(x) = \frac{d}{dx} (1 + 2x + 3x^2)^{10}$$ Applying the chain rule, we get: $$f'(x) = 10(1 + 2x + 3x^2)^9 \cdot \frac{d}{dx}(1 + 2x + 3x^2)$$ $$f'(x) = 10(1 + 2x + 3x^2)^9 (2 + 6x)$$ Now, evaluate $f'(x)$ at $x=0$: $$f'(0) = 10(1 + 2(0) + 3(0)^2)^9 (2 + 6(0))$$ $$f'(0) = 10(1)^9 (2)$$ $$f'(0) = 10 \cdot 1 \cdot 2 = 20$$ Thus, $a_1 = \frac{20}{1!} = 20$. Step 4: Conclusion. Both methods confirm that the coefficient $a_1$ is $20$.