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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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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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samdighe
VeritasPrepKarishma
bschoolaspirant9
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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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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RustyR
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.


For more check Formulas for Consecutive, Even, Odd Integers

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