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Sum of n positive integers formula for consecutive integers?

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

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New post Updated on: 20 Feb 2019, 03:34
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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.


Originally posted by RustyR on 16 Nov 2013, 17:15.
Last edited by Bunuel on 20 Feb 2019, 03:34, edited 1 time in total.
Updated.
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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New post 17 Nov 2013, 20:25
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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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New post 17 Nov 2013, 01:27
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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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New post 19 Jun 2015, 08:13
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? :roll:
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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New post 21 Jun 2015, 21:57
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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? :roll:


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

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New post 24 Jun 2015, 04:57
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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 = (300-100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers)

Sum = 200*101=20200

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

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New post 19 Dec 2018, 23:54
for consecutive odd or even ranges

sum = (f + n-1) n

f is the first number in the sequence, n is the number of elements in the sequence
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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New post 21 Dec 2018, 01:01
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Hi mathsnoob,

You are correct that the formula that you listed can be used for adding consecutive even and odd integers. In fact, it can be derived from the general approach. For example, n would be the number of terms in the sequence. The last term would be equal to f + 2(n-1), the common difference here is 2 for consecutive even and odd integers. This means the average of the first and last term is equal to [f + f + 2(n-1)]/2, which simplifies to f + (n-1). And finally we have an expression for the sum of n consecutive integers n[f + n - 1].

The problem with these highly specialized formulas is the burden of remembering the specific formula and keeping track of what each term stands for. Personally, I prefer to stick to the general formula that I listed earlier and then adapt it to a specific situation. This helps reduce the memorization burden on my brain. But I can see others who might prefer to memorize formulas for specific situations. Take your pick.

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

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New post 20 Feb 2019, 03:34
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Re: Sum of n positive integers formula for consecutive integers?   [#permalink] 20 Feb 2019, 03:34
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