Formula for consecutive, even, odd integers

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California4jx Master | Next Rank: 500 Posts Default Avatar
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Formula for consecutive, even, odd integers

Post Wed Aug 27, 2008 9:44 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Thought to put together all formulas at one place:


    Sum of all consecutive integers:

    N (N+1) / 2 ------- (I)
    where N is the number of terms.

    Example: 1+2+3+........ + 100
    N=100
    (I) => 100 (101) / 2
    => 50 (101) = 5050

    -----------------------------------------------------------------

    SUM OF EVEN NUMBERS:

    Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    Example: 2+4+6+ ....... 100
    N = (2+100)/2 - 1 = 50

    Sum of first 50 positive even integers = (50)(51) = 2550
    ------------------------------------------------------------------

    SUM OF ODD NUMBERS:

    If N = number of odd terms then sum = (N)^2

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    rishi235 Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Aug 27, 2008 9:53 am
    Nice work...
    Just to add to ur collection:

    Sum of squares of 1st n consecutive natural nos = n(n+1)(2n+1) / 6

    Sum of cubes of the 1st n consecutive natural nos. = [n(n+1)/2]^2

    California4jx Master | Next Rank: 500 Posts Default Avatar
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    Post Wed Aug 27, 2008 1:09 pm
    Great .. !

    rixyroxy Newbie | Next Rank: 10 Posts Default Avatar
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    Post Fri Aug 19, 2011 10:01 am
    Are these formulae tested on GMAT ?
    rishi235 wrote:
    Nice work...
    Just to add to ur collection:

    Sum of squares of 1st n consecutive natural nos = n(n+1)(2n+1) / 6

    Sum of cubes of the 1st n consecutive natural nos. = [n(n+1)/2]^2

    kapilhede17 Junior | Next Rank: 30 Posts Default Avatar
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    Post Fri Jul 13, 2012 9:03 pm
    I just remember the generic formula

    Sn = n/2( 2a + (n-1)d)
    a=1st term
    d= common diff

    It saves me from remembering the formula , and all the ones mentioned are like a substitution away

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    nthiru100 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Mon Nov 05, 2012 8:07 am
    Thank You Guys

    Mystiqan Newbie | Next Rank: 10 Posts
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    Post Sat Jun 21, 2014 9:40 am
    Or for even:[x(x+2)]/4; and odd:[x(x+2)+1]/4...x is the greatest digit of your set.ex: out of 7,5,3,1 your x would be 7. Hope this helps. I take credit for figuring these out without any help other than referencing the first equation.

    kiranrai.ch Newbie | Next Rank: 10 Posts
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    Post Sun Jan 10, 2016 9:08 pm
    sum of even number: n((first even+last even)/2)

    where, n = total number of even numbers

    But, you can't use the Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    For example, let's find the sum of even numbers between 102 to 200.

    =50((102+200)/2)
    =7550, which is right.

    chetan.sharma Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Jan 15, 2016 5:59 am
    kiranrai.ch wrote:
    sum of even number: n((first even+last even)/2)

    where, n = total number of even numbers

    But, you can't use the Formula: N(N+1)
    How to Find N = (First Even + Last Even)/2 - 1

    For example, let's find the sum of even numbers between 102 to 200.

    =50((102+200)/2)
    =7550, which is right.
    hi,
    the formula n(n+1) is for first n even positive integer..
    and will not be true for a set of consecutive even integers picked up from between..

    GMAT/MBA Expert

    Post Fri Feb 19, 2016 2:20 pm
    An easy way to remember the sum of consecutive integers:

    Suppose our set is [n - x, n - (x - 1), n - (x - 2), ...., n, .... n + (x - 2), n + (x - 1), n + x].

    Now let's add terms that are equidistant from the mean:

    (n - x) + (n + x) = 2n

    (n - (x - 1)) + (n + (x - 1)) = 2n

    etc.

    Wow! So every pair of terms that is equidistant from the mean has a sum of 2n, and an average of n.

    Since the sum of any set = (# of terms) * average, we know that our sum = (# of terms) * n.

    From there, just count the number of terms in the set, and you're done.

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