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Re: Sum of n positive integers formula for consecutive integers? [#permalink]
17 Nov 2013, 19:25

3

This post received KUDOS

Expert's post

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.

Re: Sum of n positive integers formula for consecutive integers? [#permalink]
17 Nov 2013, 21:29

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.

Re: Sum of n positive integers formula for consecutive integers? [#permalink]
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.

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?

Re: Sum of n positive integers formula for consecutive integers? [#permalink]
21 Jun 2015, 20:57

Expert's post

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.

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 all-applicable 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. _________________

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...