Question

The sum of odd integers from 1 to 2001 is:

• A. \( 1001^2 \)
• B. \( 1000^2 \)
• C. \( 1002^2 \)
• D. \( 1003^2 \)
• E. \( 999^2 \)

Hint:

Number of odd terms from \(1\) to \(L\) is \( \frac{L+1}{2} \) (when \(L\) is odd).

Answer 1

The Correct Option is A

Concept: The sum of first \( n \) odd natural numbers is: \[ \text{Sum} = n^2 \] So first we need to find how many odd numbers are there from \( 1 \) to \( 2001 \).

Step 1: Find the number of terms \( n \).
The sequence \( 1, 3, 5, \dots, 2001 \) is an A.P. with: \[ a = 1,\quad d = 2,\quad l = 2001 \] Using: \[ l = a + (n-1)d \] \[ 2001 = 1 + (n-1)2 \] \[ 2000 = 2(n-1) \] \[ 1000 = n - 1 \] \[ n = 1001 \]

Step 2: Apply sum formula.
\[ \text{Sum} = n^2 = (1001)^2 \]

Step 3: Expand the square.
\[ (1001)^2 = (1000 + 1)^2 \] \[ = 1000^2 + 2(1000)(1) + 1^2 \] \[ = 1000000 + 2000 + 1 = 1002001 \]

Step 4: Final answer.
\[ \boxed{1002001} \]