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Thread: formula for sum of a series

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    formula for sum of a series

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    is there a formula for sum of a series (i.e. 1+3+5+7...991)

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    ok, by pluggin in a few numbers, here's what i could come up with..let me know if it is wrong

    n/2(2f+(n-1)d)

    where n: # of terms
    f: first term in sequence
    d: difference between consecutive terms

    arriving at number of terms in sequence has a formula that i knew of..
    n = (l - f)/d
    where n: num terms
    l: last
    f: first
    d: difference

  3. #3
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    yes. much easier formula
    step 1. realize that the average of 1, 3, 5 (which is 3) is the same as the average of 1,2,3,4,5. (which is 3). in other words, as long as i'm sequential, it doesn't make a difference if i sip every other number. You can "visualize this by seeing how every number to the left of the middle gets "canceled" by a number to the right - 2 is canceled by 4 and 1 is canceled by 5. In other words, since just like 2 is "3-1" you have 4 being "3+1", therefore when you average them, you get "(2+4)/2" = "( [3-1] + [3+1])/2 = 3 so the "+1" and the "-1" cancel.
    2. know that the average of any sequence (where the numbers increase by 1) is simply the average of the first and the last number
    e.g. the average of (5,6,7,8,....195) is simply (5+195)/2.
    therefore, the average of (1,2,3,4,5,991) is = (1+991)/2 = 496
    3. since we established in part 1 that the average of (1,2,3,4,5,991) is gong to be the same as the average of (1,3,5,991), and we showed in part to that the average of (1,2,3,4,5,991) is 496, therefore the average of (1,3,5,991) is 496

    4. since Average = sum/count, therefore sum = average * count
    5. we know that the average of (1,3,5,991) is 496 (see step 3)
    6. the count of (1,3,5,991) is a bit trickier:
    since odd and even numbers go in pairs, lets first figure out how many numbers there are from 1 to 990.
    count of a sequential series (when incremented by 1) = last number - first number + 1
    thus, count of all the numbers from 1 to 990 = 990-1+1 = 990.
    since there are half as many odd numbers, there are going to be 990/2 = 495 odd numbers from 1 to 990, which means there is going to be one more odd number from 1 to 991.
    Therefore, there are going to be 496 odd numbers from 1 to 991.
    7 since sum = average * count (step 4)
    and since average = 496 (step 3)
    and since count = 496 (step 6)
    the sum = 496*496 = 246016
    i verified this number in excel.

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    Eager! WordNeard's Avatar
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    (please see above post)
    the formula is
    sum = average * count of odd numbers
    average = (first + last)/2 [not always, so know when to use this. here we can use it because the numbers are sequential by +x]
    count of odd numbers = RoundUp( [last number - first number + 1]/2)

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    An Urch Guru Pundit Swami Sage DWarrior's Avatar
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    I think the sum of n odds is equal to n^2. Probably doesn't help if you've never dealt with series before, but it is one of the easier ones. I guess that isn't entirely helpful unless you also know/can figure out the nth odd, which is just 2n-1

    Here I'd just solve 2n-1=991 => n=992/2=496
    n^2=496^2=246016

    If you practice with sequences and series like these it'll only take plugging in a couple of numbers to remember the pattern, and they're good practice for mathematical induction if you care about math. If you want to keep this one gangster, show 1=1^2 then the induction step would be assume sum of n odds = n^2 => sum of n odds + (2n-1)+2 = n^2+(2n-1)+2 => sum of n+1 odds = (n+1)^2 QED
    If you have questions about my solutions, PM me.

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    The formulae is (An-A1)/2 *n.

    1+3+5+7...991
    Here we need to find n first.
    An = A1 + (n-1)d

    991=1+ (n+1)2

    990 = 2n+2
    992 = 2n
    n=496 terms.

    Now Sum = (991 + 1)* 496/2
    = 992*496 /2
    = 496*496
    = 246016

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