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OG Problem #157

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by GHong14 » Fri Nov 26, 2010 7:31 pm
For any positive integer n, the sum of the first n positive integers equals (n(n+1))/2. what is teh sum of all the even integers between 99 and 301.

The correct answer in the OG is 20,200. The OG solves this problem by plugging 159 and 49 as n because " the sum of the even integers between 99 and 301 is the sum of the even integers from 100 through 300 or the sum of the 50th even integer through the 150 even integers. However, when solving the problem they multiply each of the equation by two: 2((150(150+1))/2) - 2((49(49+1)/2).

For the life of me i can't understand why they multiply the two formula by two. Can anyone explain?
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by Rahul@gurome » Fri Nov 26, 2010 7:56 pm
The sum of all even integers between 99 and 301 is 100 + 102 + 104…..+300 = 2(50+51+52+…+150)
= 2*(1+2+…..+150) – 2*(1+2+…+49)
=2*150*(151)/2 – 2*49*(50)/2 = 20,200.
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by GHong14 » Fri Nov 26, 2010 8:16 pm
I get the first sentence completely but can you explain the next two expressions?

= 2*(1+2+..+150) - 2*(1+2+.....+49) why did you multiply the two expression by 2
=2*150*(151)/2 - 2*49*(50)/2 = 20,200. and here divide by 2?
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by rishab1988 » Sat Nov 27, 2010 11:03 am
Here is my approach

Avg = Sum/no of terms

Therefore, sum = Avg * no of terms

For no of terms:

First term =100 [first even integer]
Last term =300 [last even integer]

No of terms = (300-100)/2 [for the even integers are spaced 2 apart]
= (200)/2 =100

Add 1 to this for the terms we are talking about are 100 to 300 inclusive

No of terms = 100+1 = 101

Average in Any Arithmetic progression [ a sequence obtained by adding a fixed constant to the previous term] :

(First term + Last term)/2 = (100+300)/2 = 200

Sum = Avg * No of terms =101* 200= 200 (100+1) =200*100 +200 = 20000+200 = 20200
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by rishab1988 » Sat Nov 27, 2010 11:09 am
As to explanation for Rahul's solution in second line.

The sentence 2*[1+2+.....+150] - 2[1+2+3..+ 49] basically is doing the simpl function of subtracting the sum of all even integers from 2 to 98 from the sum of all even integers from 2 to 300.[ this is basically a tweak to fit in the formula n(n+1)/2]

Subtracting,we basically get the sum of all integers from 100 to 300.

I hope this clears your doubt
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by GHong14 » Sat Nov 27, 2010 11:13 am
Thank you for your help! I understand it now. Thanks!!!!
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