Worksheet featuring exercises on binomial expansion, including expanding expressions, finding coefficients, and identifying terms.

Problem Overview:

The worksheet involves binomial expansions and related problems. We will solve the tasks step by step using the Binomial Theorem:

The Binomial Theorem states:
\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]
where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient,
- \( a \) and \( b \) are terms in the binomial expression,
- \( n \) is the power to which the binomial is raised.

---

Part 1: Expand Completely

#### 1. \( (1 + 2a)^5 \)
Using the Binomial Theorem:
\[
(1 + 2a)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (2a)^k
\]
Expanding term by term:
\[
\begin{aligned}
&\binom{5}{0} (1)^5 (2a)^0 + \binom{5}{1} (1)^4 (2a)^1 + \binom{5}{2} (1)^3 (2a)^2 + \binom{5}{3} (1)^2 (2a)^3 + \binom{5}{4} (1)^1 (2a)^4 + \binom{5}{5} (1)^0 (2a)^5 \\
&= 1 \cdot 1 \cdot 1 + 5 \cdot 1 \cdot 2a + 10 \cdot 1 \cdot 4a^2 + 10 \cdot 1 \cdot 8a^3 + 5 \cdot 1 \cdot 16a^4 + 1 \cdot 1 \cdot 32a^5 \\
&= 1 + 10a + 40a^2 + 80a^3 + 80a^4 + 32a^5
\end{aligned}
\]
Thus, the expansion is:
\[
\boxed{1 + 10a + 40a^2 + 80a^3 + 80a^4 + 32a^5}
\]

#### 2. \( (5b + 1)^3 \)
Using the Binomial Theorem:
\[
(5b + 1)^3 = \sum_{k=0}^{3} \binom{3}{k} (5b)^{3-k} (1)^k
\]
Expanding term by term:
\[
\begin{aligned}
&\binom{3}{0} (5b)^3 (1)^0 + \binom{3}{1} (5b)^2 (1)^1 + \binom{3}{2} (5b)^1 (1)^2 + \binom{3}{3} (5b)^0 (1)^3 \\
&= 1 \cdot 125b^3 \cdot 1 + 3 \cdot 25b^2 \cdot 1 + 3 \cdot 5b \cdot 1 + 1 \cdot 1 \cdot 1 \\
&= 125b^3 + 75b^2 + 15b + 1
\end{aligned}
\]
Thus, the expansion is:
\[
\boxed{125b^3 + 75b^2 + 15b + 1}
\]

#### 3. \( (2b - 1)^3 \)
Using the Binomial Theorem:
\[
(2b - 1)^3 = \sum_{k=0}^{3} \binom{3}{k} (2b)^{3-k} (-1)^k
\]
Expanding term by term:
\[
\begin{aligned}
&\binom{3}{0} (2b)^3 (-1)^0 + \binom{3}{1} (2b)^2 (-1)^1 + \binom{3}{2} (2b)^1 (-1)^2 + \binom{3}{3} (2b)^0 (-1)^3 \\
&= 1 \cdot 8b^3 \cdot 1 + 3 \cdot 4b^2 \cdot (-1) + 3 \cdot 2b \cdot 1 + 1 \cdot 1 \cdot (-1) \\
&= 8b^3 - 12b^2 + 6b - 1
\end{aligned}
\]
Thus, the expansion is:
\[
\boxed{8b^3 - 12b^2 + 6b - 1}
\]

#### 4. \( (3u + 1)^5 \)
Using the Binomial Theorem:
\[
(3u + 1)^5 = \sum_{k=0}^{5} \binom{5}{k} (3u)^{5-k} (1)^k
\]
Expanding term by term:
\[
\begin{aligned}
&\binom{5}{0} (3u)^5 (1)^0 + \binom{5}{1} (3u)^4 (1)^1 + \binom{5}{2} (3u)^3 (1)^2 + \binom{5}{3} (3u)^2 (1)^3 + \binom{5}{4} (3u)^1 (1)^4 + \binom{5}{5} (3u)^0 (1)^5 \\
&= 1 \cdot 243u^5 \cdot 1 + 5 \cdot 81u^4 \cdot 1 + 10 \cdot 27u^3 \cdot 1 + 10 \cdot 9u^2 \cdot 1 + 5 \cdot 3u \cdot 1 + 1 \cdot 1 \cdot 1 \\
&= 243u^5 + 405u^4 + 270u^3 + 90u^2 + 15u + 1
\end{aligned}
\]
Thus, the expansion is:
\[
\boxed{243u^5 + 405u^4 + 270u^3 + 90u^2 + 15u + 1}
\]

---

Part 2: Find Each Coefficient Described

#### 11. Coefficient of \( x^{20} \) in expansion of \( (1 - 2x^4)^7 \)
The general term in the expansion is:
\[
T_k = \binom{7}{k} (1)^{7-k} (-2x^4)^k = \binom{7}{k} (-2)^k x^{4k}
\]
We need the term where the exponent of \( x \) is 20:
\[
4k = 20 \implies k = 5
\]
Substitute \( k = 5 \) into the general term:
\[
T_5 = \binom{7}{5} (-2)^5 x^{20}
\]
Calculate the coefficient:
\[
\binom{7}{5} = \binom{7}{2} = \frac{7 \cdot 6}{2 \cdot 1} = 21
\]
\[
(-2)^5 = -32
\]
Thus, the coefficient is:
\[
21 \cdot (-32) = -672
\]
The coefficient of \( x^{20} \) is:
\[
\boxed{-672}
\]

#### 12. Coefficient of \( y^4 x^2 \) in expansion of \( (2y - 3x^2)^5 \)
The general term in the expansion is:
\[
T_k = \binom{5}{k} (2y)^{5-k} (-3x^2)^k = \binom{5}{k} 2^{5-k} (-3)^k y^{5-k} x^{2k}
\]
We need the term where the exponents are \( y^4 \) and \( x^2 \):
\[
5 - k = 4 \implies k = 1
\]
Substitute \( k = 1 \) into the general term:
\[
T_1 = \binom{5}{1} 2^{5-1} (-3)^1 y^4 x^{2 \cdot 1}
\]
Calculate the coefficient:
\[
\binom{5}{1} = 5
\]
\[
2^{5-1} = 2^4 = 16
\]
\[
(-3)^1 = -3
\]
Thus, the coefficient is:
\[
5 \cdot 16 \cdot (-3) = -240
\]
The coefficient of \( y^4 x^2 \) is:
\[
\boxed{-240}
\]

---

Part 3: Find Each Term Described

#### 21. 4th term in expansion of \( (1 - 5x^3)^3 \)
The general term in the expansion is:
\[
T_k = \binom{3}{k} (1)^{3-k} (-5x^3)^k = \binom{3}{k} (-5)^k x^{3k}
\]
The 4th term corresponds to \( k = 3 \):
\[
T_3 = \binom{3}{3} (-5)^3 x^{3 \cdot 3}
\]
Calculate the term:
\[
\binom{3}{3} = 1
\]
\[
(-5)^3 = -125
\]
\[
x^{3 \cdot 3} = x^9
\]
Thus, the 4th term is:
\[
1 \cdot (-125) \cdot x^9 = -125x^9
\]
The 4th term is:
\[
\boxed{-125x^9}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&1. \quad 1 + 10a + 40a^2 + 80a^3 + 80a^4 + 32a^5 \\
&2. \quad 125b^3 + 75b^2 + 15b + 1 \\
&3. \quad 8b^3 - 12b^2 + 6b - 1 \\
&4. \quad 243u^5 + 405u^4 + 270u^3 + 90u^2 + 15u + 1 \\
&11. \quad -672 \\
&12. \quad -240 \\
&21. \quad -125x^9
\end{aligned}
}
\]