(i) Find the sum of odd integers from 1to 2001.
(ii) Find the sum of all natural numbers between 99 and 1001, which are multiples of 5.

Answer

Let's break down the solution to the question step by step.

### Part (i): Find the sum of odd integers from 1 to 2001.

1. **Identify the series**: The series of odd integers from 1 to 2001 is:
\[ S_n = 1 + 3 + 5 + \ldots + 2001 \]

2. **Determine the number of terms (n)**: The nth term of an arithmetic series can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a \) is the first term (1), \( d \) is the common difference (2), and \( a_n \) is the last term (2001). Setting up the equation:
\[ 2001 = 1 + (n-1) \cdot 2 \]
Rearranging gives:
\[ 2001 - 1 = (n-1) \cdot 2 \implies 2000 = (n-1) \cdot 2 \implies n-1 = 1000 \implies n = 1001 \]

3. **Use the formula for the sum of an arithmetic series**: The sum \( S_n \) can be calculated using the formula:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Substituting the values \( n = 1001 \), \( a = 1 \), and \( d = 2 \):
\[ S_n = \frac{1001}{2} \cdot (2 \cdot 1 + (1001 - 1) \cdot 2) \]
Simplifying:
\[ S_n = \frac{1001}{2} \cdot (2 + 2000) = \frac{1001}{2} \cdot 2002 \]
\[ S_n = 1001 \cdot 1001 = 1002001 \]

### Answer for Part (i):
The sum of odd integers from 1 to 2001 is **1002001**.

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### Part (ii): Find the sum of all natural numbers between 99 and 1001, which are multiples of 5.

1. **Identify the series**: The multiples of 5 between 99 and 1001 start from 100 and go up to 1000:
\[ S_n = 100 + 105 + 110 + \ldots + 1000 \]

2. **Determine the number of terms (n)**: The nth term can be expressed as:
\[ a_n = a + (n-1)d \]
where \( a = 100 \), \( d = 5 \), and \( a_n = 1000 \):
\[ 1000 = 100 + (n-1) \cdot 5 \]
Rearranging gives:
\[ 1000 - 100 = (n-1) \cdot 5 \implies 900 = (n-1) \cdot 5 \implies n-1 = 180 \implies n = 181 \]

3. **Use the formula for the sum of an arithmetic series**: The sum \( S_n \) can be calculated using:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
Substituting the values \( n = 181 \), \( a = 100 \), and \( d = 5 \):
\[ S_n = \frac{181}{2} \cdot (2 \cdot 100 + (181 - 1) \cdot 5) \]
Simplifying:
\[ S_n = \frac{181}{2} \cdot (200 + 900) = \frac{181}{2} \cdot 1100 \]
\[ S_n = 181 \cdot 550 = 99550 \]

### Answer for Part (ii):
The sum of all natural numbers between 99 and 1001 that are multiples of 5 is **99550**.

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