Sum of all even natural numbers between 1 and 2001?

Question

What is the sum of all the even natural numbers between 1 and 2001?

Answer 1

There is a 1:1 correspondence between the terms of 2, 4, 6 ... 2000 and the terms of 1, 2, 3 ,,, 1000 and the sum of the former is twice the sum of the latter;

Pairing terms as follows in the latter (1+1000) + (2 + 999) + (3 + 998) etc we can see the sum of 1, 2, 3 ,,, 1000 is 500 x 1001 = 500500

and then we double it to get the sum of 2, 4, 6 ... 2000 = 1001000

Answer 2

e^x gave you the general formula to answer your type of question.

another approach, that may help you understand the problem and why e^s's formula is correct, is to note that there are 1000 even numbers between 1 and 2001, namely 2, 4, 6, through 1998, 2000.

these form 500 pairs of numbers where each pair adds up to 2002, namely 2+2000, 4+1998, 6 + 1996, through 998 + 1004, 1000 +1002.

multiply 500 x 2002 for your answer: 1,001,000

good luck!

Answer 3

sum = 2 + 4 + ... + 1998 + 2000
sum/2 = 1 + 2 + ... + 999 + 1000 =
sum/2 = (1 + 1000) + (2 + 999) + ... . = 1001 + 1001 + ... .
sum/2 = 500 x 1001 = 500500
sum = 100100

Answer 4

e^x is right with the arithmetic sequence. if you're adding even numbers starting with 2, an easier way would be to follow the following steps:

example: sum of even numbers from 2 to 2001

1. add the last number to 2

2 + 2000 = 2002

2. divide the result by 2

2002 / 2 = 1001

3. subtract 1 from the result in (#2)

1001 - 1 = 1000

4. multiply (#2) and (#3)

1001 * 1000 = 1,001,000

hope this one helps! =)

Answer 5

This is simply the sum of an arithmetic sequence with 1000 terms, the first term being 2 and the common difference being 2.

This is S= (n/2)(2a + (n-1)d)

n = terms = 1000
a = first term = 2
d = difference = 2

S = (500)(4 + 2(999))

or 1001000

Answer 6

sum=n(n+1)=1000(1000+1)=1001000
where n=the number of even natural numbers

Answer 7

2+4+6+8+...2000 = (2)(1+2+3+...+1000)

1+2+3+...+1000 = (1000)(1000+1)/2 = 500500

(2)(500500) = 1001000

Answer 8

1001000 as explained by others