Let's solve this step by step.

Part (i): Binomial Expansion of $(2 + x)^5$

The binomial expansion formula is:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here:

• $a = 2$, $b = x$, and $n = 5$.

We expand $(2 + x)^5$ term by term:

$(2 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} x^k$

Now calculate each term where $k = 0, 1, 2, 3, 4, 5$:

1. For $k = 0$: $\binom{5}{0} 2^{5-0} x^0 = 1 \cdot 2^5 \cdot 1 = 32$

2. For $k = 1$: $\binom{5}{1} 2^{5-1} x^1 = 5 \cdot 2^4 \cdot x = 5 \cdot 16 \cdot x = 80x$

3. For $k = 2$: $\binom{5}{2} 2^{5-2} x^2 = 10 \cdot 2^3 \cdot x^2 = 10 \cdot 8 \cdot x^2 = 80x^2$

4. For $k = 3$: $\binom{5}{3} 2^{5-3} x^3 = 10 \cdot 2^2 \cdot x^3 = 10 \cdot 4 \cdot x^3 = 40x^3$

5. For $k = 4$: $\binom{5}{4} 2^{5-4} x^4 = 5 \cdot 2^1 \cdot x^4 = 5 \cdot 2 \cdot x^4 = 10x^4$

6. For $k = 5$: $\binom{5}{5} 2^{5-5} x^5 = 1 \cdot 1 \cdot x^5 = x^5$

Final Expansion:

$(2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$

Part (ii): Coefficient of $y^3$ in $(2 + 3y + y^2)^5$

Here, we are tasked to find the coefficient of $y^3$ in the expansion of $(2 + 3y + y^2)^5$.

To solve this, we apply the multinomial expansion formula:

$(a + b + c)^n = \sum_{i+j+k=n} \frac{n!}{i!j!k!} a^i b^j c^k$

Here:

• $a = 2$, $b = 3y$, $c = y^2$, and $n = 5$.
• We need to find the terms where the total power of $y$ is 3.

Step 1: Identify Possible Combinations

The power of $y$ comes from:

• $b^j = (3y)^j$ contributes $j$ powers of $y$,
• $c^k = (y^2)^k$ contributes $2k$ powers of $y$.

We need $j + 2k = 3$, where $i + j + k = 5$. Let's check possible values of $j$ and $k$:

1. $j = 3, k = 0$ → $j + 2k = 3$, valid.
2. $j = 1, k = 1$ → $j + 2k = 3$, valid.

Thus, there are two contributing terms.

Step 2: Compute Each Term

1. Case 1: $j = 3, k = 0$

   • Here $i = 5 - j - k = 5 - 3 - 0 = 2$.
   • The term is: $\frac{5!}{2!3!0!} (2)^2 (3y)^3 (y^2)^0$ Simplify: $\frac{120}{2 \cdot 6} \cdot 4 \cdot 27y^3 \cdot 1 = 10 \cdot 4 \cdot 27y^3 = 1080y^3$

2. Case 2: $j = 1, k = 1$

   • Here $i = 5 - j - k = 5 - 1 - 1 = 3$.
   • The term is: $\frac{5!}{3!1!1!} (2)^3 (3y)^1 (y^2)^1$ Simplify: $\frac{120}{6 \cdot 1 \cdot 1} \cdot 8 \cdot 3y \cdot y^2 = 20 \cdot 8 \cdot 3y^3 = 480y^3$

Step 3: Add Coefficients

The total coefficient of $y^3$ is:

$1080 + 480 = 1560$

Final Answer:

1. Part (i): $(2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5$
2. Part (ii): The coefficient of $y^3$ in the expansion of $(2 + 3y + y^2)^5$ is 1560.

Do you need any further clarifications or explanations? Let me know!

Related Questions:

1. How do you generalize the multinomial theorem for three variables?
2. How do you find the coefficient of specific terms in multinomial expansions?
3. Can you derive the binomial theorem for negative or fractional powers?
4. What happens when $n$ in the binomial theorem is very large?
5. How can Pascal’s triangle help in expanding binomials?

Tip:

When solving multinomial problems, ensure the exponents add up to the total power $n$, and systematically check all combinations.