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Sum of n positive integers formula for consecutive integers? [#permalink]
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16 Nov 2013, 16:15
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Is the sum of n positive integers formula, \(\frac{n(n+1)}{2}\), applicable to sets of consecutive integers as well? For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x \(\frac{20(21)}{2}\) In OG PS Q172, a similar approach has been used for even numbers.



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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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17 Nov 2013, 00:27
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3+6+9+12+15+18+21+24+27+30...... = 3*(1+2+3+4+5+6+7+8+9+10...) So, yup.. You can do that...
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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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bschoolaspirant9 wrote: Is the sum of n positive integers formula, \(\frac{n(n+1)}{2}\), applicable to sets of consecutive integers as well? For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x \(\frac{20(21)}{2}\) In OG PS Q172, a similar approach has been used for even numbers. Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1. A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula. In the even integers questions, you may be required to find the sum of first 10 even integers. 2 + 4 + 6 + ... + 18 + 20 Take 2 common, 2*( 1 + 2 + 3 + ...10)To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum. Note that for odd integers, you cannot directly use this formula. Sum the first 10 odd integers 1 + 3 + 5 + 7+...+19 But you can still make some modifications to find the sum. 1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20)  (2 + 4+ 6+...20) We know how to sum consecutive integers. (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2 (2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before) So 1 + 3 + 5 + 7+...+19 = (20*21/2)  (10*11) = 100 The direct formula of sum of n consecutive odd integers starting from 1 = n^2
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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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17 Nov 2013, 21:29
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VeritasPrepKarishma wrote: bschoolaspirant9 wrote: Is the sum of n positive integers formula, \(\frac{n(n+1)}{2}\), applicable to sets of consecutive integers as well? For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x \(\frac{20(21)}{2}\) In OG PS Q172, a similar approach has been used for even numbers. Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1. A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula. In the even integers questions, you may be required to find the sum of first 10 even integers. 2 + 4 + 6 + ... + 18 + 20 Take 2 common, 2*( 1 + 2 + 3 + ...10)To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum. Note that for odd integers, you cannot directly use this formula. Sum the first 10 odd integers 1 + 3 + 5 + 7+...+19 But you can still make some modifications to find the sum. 1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20)  (2 + 4+ 6+...20) We know how to sum consecutive integers. (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2 (2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before) So 1 + 3 + 5 + 7+...+19 = (20*21/2)  (10*11) = 100 The direct formula of sum of n consecutive odd integers starting from 1 = n^2 Thank you for the detailed response.



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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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11 Jan 2015, 10:17
Great help guys, an Karishma, huge contribution, thanks from Chile!



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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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19 Jun 2015, 07:13
VeritasPrepKarishma wrote: bschoolaspirant9 wrote: Is the sum of n positive integers formula, \(\frac{n(n+1)}{2}\), applicable to sets of consecutive integers as well? For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x \(\frac{20(21)}{2}\) In OG PS Q172, a similar approach has been used for even numbers. Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1. A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula. In the even integers questions, you may be required to find the sum of first 10 even integers. 2 + 4 + 6 + ... + 18 + 20 Take 2 common, 2*( 1 + 2 + 3 + ...10)To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum. Note that for odd integers, you cannot directly use this formula. Sum the first 10 odd integers 1 + 3 + 5 + 7+...+19 But you can still make some modifications to find the sum. 1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20)  (2 + 4+ 6+...20) We know how to sum consecutive integers. (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2 (2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before) So 1 + 3 + 5 + 7+...+19 = (20*21/2)  (10*11) = 100 The direct formula of sum of n consecutive odd integers starting from 1 = n^2 Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?
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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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21 Jun 2015, 20:57
samdighe wrote: VeritasPrepKarishma wrote: bschoolaspirant9 wrote: Is the sum of n positive integers formula, \(\frac{n(n+1)}{2}\), applicable to sets of consecutive integers as well? For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x \(\frac{20(21)}{2}\) In OG PS Q172, a similar approach has been used for even numbers. Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1. A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula. In the even integers questions, you may be required to find the sum of first 10 even integers. 2 + 4 + 6 + ... + 18 + 20 Take 2 common, 2*( 1 + 2 + 3 + ...10)To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum. Note that for odd integers, you cannot directly use this formula. Sum the first 10 odd integers 1 + 3 + 5 + 7+...+19 But you can still make some modifications to find the sum. 1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20)  (2 + 4+ 6+...20) We know how to sum consecutive integers. (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2 (2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before) So 1 + 3 + 5 + 7+...+19 = (20*21/2)  (10*11) = 100 The direct formula of sum of n consecutive odd integers starting from 1 = n^2 Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct? We can, provided we know the average and the number of terms. If we are asked to find the sum of first 50 consecutive positive odd integers, it might be easier to use 50^2 than to find average and then find the sum. Mind you, I myself believe in knowing just the main allapplicable kind of formulas and then twisting them around to apply to any situation. But some people prefer to work more on specific formulas and these discussions are for their benefit.
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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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24 Jun 2015, 03:57
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The best way to deal with sum of any type of arithmetic series is to use the following expression: Sum = Average*Number of terms Average = (First term + Last term)/2 Note: The average or the median can be related to the first and last terms, which are often given for these types of problems. Number of Terms = (Last Term  First Term)/Spacing + 1 Also, I recommend that you understand the basis of these relationships as opposed to blindly memorizing them. The above relationships are flexible enough to allow you to deal with any GMAT problem on these concepts. For example in Q172 from the Official Guide: What is the sum of all the even integers between 99 and 301 ? Average = (100+300)/2 = 200 Number of Terms = (300100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers) Sum = 200*101=20200 Cheers, Dabral



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