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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] New post 16 Nov 2013, 16:15
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] New post 17 Nov 2013, 00:27
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] New post 17 Nov 2013, 19:25
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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] New post 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.

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