Display Problems

Imagination is more important than knowledge... Albert Einstein

Guess is more important than calculation --- Knowhowacademy.com

Sequence - Sum

SOURCE:COMPETITION
Number of Problems: 4.
FOR PRINT ::: (Book)

Problem Num : 1

Type: None

Topic:Sequence

Theme:None
Adjustment# :

Difficulty: 1

Category Sum

Analysis

Solution/Hint

Problem Num : 2

Type:

Topic:Sequence

Adjustment# : 0

Difficulty: 1

'
What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?
$mathrm{(A) } 0qquad mathrm{(B) } 1qquad mathrm{(C) } 2qquad mathrm{(D) } 2003qquad mathrm{(E) } 4006$
'

Category Sum

Analysis

Solution/Hint
Solution 1

The first $2003$ even counting numbers are $2,4,6,...,4006$ .
The first $2003$ odd counting numbers are $1,3,5,...,4005$ .
Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$ .
$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$
$= 1+1+1+...+1 = oxed{mathrm{(D)} 2003}$

Solution 2

Using the sum of an arithmetic progression formula, we can write this as $frac{2003}{2}(2 + 4006) - frac{2003}{2}(1 + 4005) = frac{2003}{2} cdot 2 = oxed{mathrm{(D)} 2003}$ .

Solution 3

The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$ , (E standing for even).
Sum of first $n$ odd numbers, is $S_O=n^{2}$ , (O standing for odd).
Knowing this, plug $2003$ for $n$ ,
$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 Rightarrow$ $oxed{mathrm{(D)} 2003}$ .

Problem Num : 3

Type: Calculation

Topic:Sequence

Theme:Manipulation
Adjustment# : 0

Difficulty: 1

Category Sum

Analysis

Solution/Hint

Problem Num : 4

Type: Complex

Topic:Sequence

Theme:Manipulation
Adjustment# : 0

Difficulty: 3

Category Sum

Analysis

Solution/Hint