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For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) What is the sum

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by Vincen » Tue Jan 19, 2021 9:47 am

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For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) What is the sum of all the even integers between \(99\) and \(301?\)

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Answer: B

Source: Official Guide
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Source: — Problem Solving |

Vincen wrote:
Tue Jan 19, 2021 9:47 am
For any positive integer \(n,\) the sum of the first \(n\) positive integers equals \(\dfrac{n(n+1)}2.\) What is the sum of all the even integers between \(99\) and \(301?\)

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Answer: B

Source: Official Guide
Solution:

The phrase “between 99 and 301” does not include the endpoints. The sum of the even integers from 100 to 300 is:

[(100 + 300)/2] * [(300 - 100)/2 + 1]

= 200 * 101

= 20,200

[Note: The first factor is the average of the even integers and the second factor is the number of even integers from 100 to 300.]

Answer: B

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