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Re: For any positive integer n, the sum of the first n positive [#permalink]
08 Dec 2013, 18:32
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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
Approach #1: Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: math-number-theory-88376.html)
Approach #2: Using the formula of the sum of the first n positive integers: n(n+1)/2.
100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.
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