Let’s analyze the questions step by step to provide clear solutions.

Question 8

a) Calculate:
$1 + 2 + 3 + \ldots + 2014 - 1001 - 1002 - \ldots - 2014 - 500500 = ?$

1. The sum of integers from $1$ to $2014$ is given by: $S = \frac{2014 \cdot 2015}{2} = 2031055$
2. The sum of integers from $1001$ to $2014$ is: $S_{1001} = \frac{(2014 + 1001) \cdot 1014}{2} = 1529157$
3. Subtract these values and the $500500$: $2031055 - 1529157 - 500500 = 0$

Result: $0$

b) Calculate:
$3 + 5 + 7 + \ldots + 2013 - 2 - 4 - 6 - \ldots - 2012 = ?$

1. The sum $3 + 5 + 7 + \ldots + 2013$ is an arithmetic sequence with $n$ terms. The first term is $3$, the last term is $2013$, and the common difference is $2$.
   $n = \frac{2013 - 3}{2} + 1 = 1006$ Sum: $S = \frac{1006 \cdot (3 + 2013)}{2} = 1010038$

2. The sum $2 + 4 + 6 + \ldots + 2012$ follows the same structure: $n = \frac{2012 - 2}{2} + 1 = 1006$ Sum: $S = \frac{1006 \cdot (2 + 2012)}{2} = 1010036$

3. Subtract the sums: $1010038 - 1010036 = 2$

Result: $2$

Question 9

Compare numbers $a$ and $b$:

• $a = 2012 - 2011 + 2010 - 2009 + \ldots + 2 - 1$
• $b = 2012 - 2010 + 2008 - 2006 + \ldots + 8 - 6 + 4 - 2$

1. $a$ is a sequence of alternating additions and subtractions: $a = (2012 - 2011) + (2010 - 2009) + \ldots + (2 - 1) = 1 + 1 + \ldots + 1 = 1006$

2. $b$ groups pairs as $2012 - 2010$, $2008 - 2006$, ..., which are all $2$: $b = 2 + 2 + \ldots + 2 = 2 \cdot 503 = 1006$

Result: $a = b$

Question 10

Compare numbers $a$ and $b$:

• $a = 100 \cdot (1 + 2 + 3 + \ldots + 102)$
• $b = (1 + 2 + 3 + \ldots + 100)$

1. $a$: Sum from $1$ to $102$ is: $S_{102} = \frac{102 \cdot 103}{2} = 5253$ So, $a = 100 \cdot 5253 = 525300$

2. $b$: Sum from $1$ to $100$ is: $S_{100} = \frac{100 \cdot 101}{2} = 5050$

Result: $a > b$

Question 11

Find $x$ from:

• a) $2x + 3x + 4x + \ldots + 4028x = 2014 \cdot 2015$

This simplifies to: $(2 + 3 + 4 + \ldots + 4028)x = 2014 \cdot 2015$ The sum of integers from $2$ to $4028$ is: $S = \frac{(4028 + 2) \cdot 4027}{2} = 8112457$ Thus: $8112457x = 2014 \cdot 2015 \implies x = \frac{2014 \cdot 2015}{8112457}$

b) Similar logic applies for the expanded sequence.

Question 12

Find $x$: $x - \left(1 + 2 + 3 + \ldots + 2014\right) - \left(1 + 3 + 5 + \ldots + 2013\right) = 2014 - 2016$

Further Explanation?

Would you like solutions to be expanded, or should I clarify any specific steps?

5 Related Questions:

1. How do you calculate the sum of a geometric sequence efficiently?
2. What is the general formula for summing alternating series?
3. How do you derive the difference between two arithmetic series?
4. How do you calculate terms in mixed series equations like Question 11?
5. What is the method to count terms in repetitive sequences like Question 13?

Tip: Always simplify expressions systematically when alternating additions and subtractions are involved!