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# For any positive integer n, the sum of the first n positive

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Intern
Joined: 10 Oct 2008
Posts: 14

Kudos [?]: 10 [0], given: 0

For any positive integer n, the sum of the first n positive [#permalink]

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17 Oct 2008, 10:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For any positive integer n, the sum of the first n positive integers equals [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

Kudos [?]: 10 [0], given: 0

SVP
Joined: 17 Jun 2008
Posts: 1534

Kudos [?]: 279 [0], given: 0

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17 Oct 2008, 12:48
B.

Required sum = 100 + 102 + 104+.....+300
= 2(50 + 51 + 52 + .......+150)
=2(sum of integers from 1 to 150 - sum of integers from 1 to 49)
=2*150*151/2 - 2*49*50/2 = 20200.

Kudos [?]: 279 [0], given: 0

Re: sum   [#permalink] 17 Oct 2008, 12:48
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