Let's tackle each part of your question step by step, showing all work and detailed explanations for better clarity.

Section 1.4 (Binary Representations)

(d) Convert the integer 100 to binary:

To find the binary representation of 100, we will repeatedly divide by 2 and keep track of the remainders. Here's how:

1. $100 \div 2 = 50$ remainder 0
2. $50 \div 2 = 25$ remainder 0
3. $25 \div 2 = 12$ remainder 1
4. $12 \div 2 = 6$ remainder 0
5. $6 \div 2 = 3$ remainder 0
6. $3 \div 2 = 1$ remainder 1
7. $1 \div 2 = 0$ remainder 1

Now, writing the remainders from bottom to top, we get: Binary of 100 = $1100100_2$

(d) Convert the binary 10011110000 to decimal:

Now, we convert the binary number $10011110000_2$ to its decimal equivalent.

The binary number $10011110000_2$ represents:

$1 \times 2^{10} + 0 \times 2^9 + 0 \times 2^8 + 1 \times 2^7 + 1 \times 2^6 + 1 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 0 \times 2^0$

Now calculate the powers of 2 and sum the results:

$1024 + 128 + 64 + 32 + 16 = 1264$

So, $10011110000_2$ in decimal is 1264.

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Section 1.5 (Series Calculation)

(d) Calculate the series $\sum_{k=1}^{n} (2k-1)$ for $n = 4$

The given series is: $\sum_{k=1}^{4} (2k - 1)$

We will calculate each term of the series:

1. For $k = 1$, $2(1) - 1 = 1$
2. For $k = 2$, $2(2) - 1 = 3$
3. For $k = 3$, $2(3) - 1 = 5$
4. For $k = 4$, $2(4) - 1 = 7$

Now sum the values: $1 + 3 + 5 + 7 = 16$

So, $\sum_{k=1}^{4} (2k - 1) = 16$.

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Section 1.6 (Set Operations)

Given the sets:

• $A_k = \{x \in \mathbb{Q} : k - 1 < x \leq k\}$
• $B_k = \{x \in \mathbb{Q} : -k < x < k\}$

Let's evaluate the union and intersection of these sets for $i = 1$ to $i = 5$.

(a) $\bigcup_{i=1}^{5} A_i$

Here, each $A_i$ represents an interval:

• $A_1 = (0, 1]$
• $A_2 = (1, 2]$
• $A_3 = (2, 3]$
• $A_4 = (3, 4]$
• $A_5 = (4, 5]$

The union of these intervals is: $\bigcup_{i=1}^{5} A_i = (0, 5]$

(b) $\bigcup_{i=1}^{5} B_i$

Here, each $B_i$ represents an interval:

• $B_1 = (-1, 1)$
• $B_2 = (-2, 2)$
• $B_3 = (-3, 3)$
• $B_4 = (-4, 4)$
• ( B_5 = (-