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Help with sum of first n integers.

  1. Oct 17, 2004 #1
    Ok, I haven't done maths for a few years now and I've been set the following question:

    The sum of the first integers is given by:

    Sum(n) = 1+2+3+4 ....... +n = n(n+1)/2

    Find similar formulae for

    Even(n) = 2+4+6+8 ....... +2n
    Odd(n) = 1+3+5+7 ........ +(2n-1)

    Now the formulae I have come up with, by trial and error mind, are:

    Even = n(n+2)/4
    Odd = n(n)/4

    Am I on the right lines here? And if so is there a mathematical way of coming up with these formuale that doesn't involve trial and error?

    cheers
     
  2. jcsd
  3. Oct 17, 2004 #2

    arildno

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    Welcome to PF!
    Question:
    Have you checked at all whether your formulae are correct??
    What is Even(1) ?
    What does your formula tell you that Even(1) should be?
     
  4. Oct 17, 2004 #3
    well my formula says even 1 should be 0.75 , but i don't understand how you can test with even 1 anyway when there aren't any even integers before 1
     
  5. Oct 17, 2004 #4

    arildno

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    The FIRST even, positive integer is 2.
    Hence, Even(1)=2.
    You are to sum the n first even integers, so n itself can perfectly well be odd..
     
  6. Oct 17, 2004 #5
    ::thinking:: How is the nth element of the series of even numbers related to the nth element of the series of all integers? Then consider how is the nth element of the series of odd numbers related to the nth element of the even numbers?
     
    Last edited by a moderator: Oct 17, 2004
  7. Oct 17, 2004 #6
    bah now im really confused :frown:

    ok i get what you're saying Even (1) refers to the first even integer. So my formula is incorrect as the answer I should expect to get is 2. now to think up a new forumla :confused:
     
  8. Oct 17, 2004 #7

    arildno

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    Let's take it easy.
    The sum of the first n (even AND odd) integers S(n) is given by:
    [tex]S(n)=\frac{n(n+1)}{2}[/tex]
    We are to find a formula for the SUM of first n even integers, Even(n).
    Written out, this is:
    Even(n)=2+4+6++++2n (n TERMS in the sum!!)
    We may draw out a common factor of 2:
    Even(n)=2(1+2+3+++n)=2S(n) agreed?
    Hence, Even(n)=n(n+1)
    Did you follow that?
    (If not, please make SPECIFIC reference to the steps you have trouble with!)
     
    Last edited: Oct 17, 2004
  9. Oct 17, 2004 #8
    yeah i follow it right up until the end where i just confuse myself.

    ok the formula for calculating the addition of n integers is:

    n(n+1)/2

    so for all the integers up to and including 10, added together, using the formula the answer is 55 which is correct.

    Now from what you were saying the formula for the Even number of n integers added together is n(n+1) but how is this so if, using 10 as an example again, the answer comes to 110
     
    Last edited: Oct 17, 2004
  10. Oct 17, 2004 #9

    arildno

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    But: Even(n) is the sum of the n first even integers.
    Up to and including "10" we have 5 even integers:2,4,6,8,10
    The sum of these are 5*(5+1)=30.
     
  11. Oct 17, 2004 #10
    aaaah ok i feel so stupid now. right, it has clicked. thank you so much, pretty sure I will be able to figure out the formula for the odd numbers now. its always the blindingly obvious i slip up on :rofl:
     
  12. Oct 17, 2004 #11

    arildno

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    You're welcome..
     
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