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Sum of First 50 Odd Numbers

  1. Jun 14, 2009 #1
    I'm working on a homework sheet that will represent where I stand mathematically in order to qualify for a summer PreCal course. I remember my Algebra 2 teacher showing me how to do this problem using a calculator, but I never learned how to do this by hand. Any help is appreciated.


    1. The problem statement, all variables and given/known data
    Find the sum of the first 50 odd numbers:

    50
    [tex]\sum\[/tex] 2n-1
    n=1


    2. Relevant equations

    Summation problem.

    3. The attempt at a solution

    As I said, I've only solved this specific type of summation problem using a calculator, but I know how to do regular summation problems, for instance:

    6
    [tex]\sum\[/tex] 2n-1
    n=1

    2(1)-1 = 1
    2(2)-1 = 3
    2(3)-1 = 5
    2(4)-1 = 7
    2(5)-1 = 9
    2(6)-1 = 11
    -----
    36

    I hope that qualifies as an attempt.
     
  2. jcsd
  3. Jun 14, 2009 #2

    Cyosis

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    Homework Helper

    Just doing the summation manually is kind of tedious. Do you know a formula for the sum of the first n natural numbers?
     
  4. Jun 14, 2009 #3
    No I don't know the formula, but I'm sure it would help :)
     
  5. Jun 14, 2009 #4

    Cyosis

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    Homework Helper

    Have you ever heard the story about Gauss adding up the first 100 natural numbers in mere seconds? I hope you have or this formula won't make much sense to you.

    [tex]
    \sum_{k=1}^n k=\frac{n}{2}(n+1)
    [/tex]

    Use this sum to calculate the sum you're interested in.
     
  6. Jun 14, 2009 #5
    Ouch, no I haven't.
     
  7. Jun 14, 2009 #6

    jgens

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    Gold Member

    Perhaps to help you derive a formula for the sum, let S be the finite sum,

    S = n0 + (n0 + d) + (n0 + 2d) + . . . + (n0 + (n-1)d)

    Where n0 represents the first term of the sum and n represents the number of terms.

    Since addition is communicative and associative, if we sum S twice we find that,

    2S = [(n0) + (n0 + (n-1)d)] + [(n0 + d) + (n0 + (n-2)d)] + . . .

    2S = (2n0 + (n-1)d) + (2n0 + (n-1)d) + . . .

    Try to finish the derivation and see how it applies to the sum you're trying to compute. Sorry if this is hard to understand.
     
  8. Jun 14, 2009 #7

    jgens

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    Gold Member

    Cyosis, why would he/she use the sum for the first n natural numbers when his sum asks for odd natural numbers only?
     
  9. Jun 14, 2009 #8

    Cyosis

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    Homework Helper

    If you know the sum for the first n natural numbers you can quickly solve the sum for the first odd numbers. Had he known it, it would have been the easiest way to calculate the sum he's interested in.
     
  10. Jun 14, 2009 #9
    The story is that a six-year-old Gauss was told by his teacher to add the numbers 1 to 100 as a form of busywork. In seconds, he came up with the answer, much to his teacher's surprise. What he did was notice that since 1+100=101, 2+99=101, 3+98=101, etc., the sum becomes a simple multiplication: there are 50 of the above "pairs," so the sum is just 50*101 = 5050. Anyway, that's the intuition behind the identity that Cyosis posted. You can pretty much use the exact same intuition.
     
  11. Jun 14, 2009 #10
    Okay thanks, I get it now.
     
  12. Jun 14, 2009 #11

    Cyosis

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    Are you able to use this information so far? Depending on where you are in your education we may want to use different methods. Is this a first course on summations, do you know what arithmetic progression is?
     
  13. Jun 14, 2009 #12
    Also, one of the things I like about Wolfram Alpha is that it tries to guess what series you're trying to input, as long as it's "simple." So, if you type in 1+3+5+..., it guesses that you're entering a sum of odd numbers, and spits out an analytical expression for the partial sum.
     
  14. Jun 14, 2009 #13
    Well, I get some of it, but I'm thinking there may be an easier way for me. I just came out of an algebra 2 course where we learned simple summations like the one I used as an example in my original post. We only did one or two of the type of summation I'm currently asking about, and it was all on the calculator with no formula, we used pre-programed apps.

    Is an arithmetic progression the same as an arithmetic sequence? I apologize if that's a stupid question.
     
  15. Jun 14, 2009 #14

    jgens

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    Gold Member

    Well, if you're interested, here's a couple of derivations for summation formulas. Sometimes knowing how formulas are derived is useful (at least I find it useful).

    Arithmetic Sums:

    Suppose we have an arithmetic sequence with a constant difference d between terms. We let the sum, SA represent the sum of the first n terms of that sequence such that

    SA= k0 + k1 + . . . + kn-1 + kn

    Where k0 represents the first term of the series, k1 = k0 + d, k2 = k1 + d, and so on and so forth. Given our definition of ki it follows that,

    SA = k0 + (k0 + d) + . . . + (k0 + (n-2)d) + (k0 + (n-1)d)

    Since addition is both communicative and associative, we can take the sum twice and rearrange the terms such that,

    2SA = (k0 + kn) + (k1 + kn-1) + . . .

    Therefore,

    2SA = [k0 +(k0 + (n-1)d)] + [(k0 + d) + (k0 + (n-2)d)] + . . .

    2SA = (2k0 + (n-1)d) + (2k0 + (n-1)d) + . . .

    Since we have n number of terms in the sum, it follows that,

    2SA = n(2k0 + (n-1)d)

    and consequently, we have that,

    SA = (n/2)(2k0 + (n-1)d)


    From this, we see that the sum of the first n natural numbers is just a specific case of an arithmetic sequence. We know that the first term, k0 = 1 and we know that d = 1. So the sum sA of the first n natural numbers is,

    sA = (n/2)(n+1)


    Geometric Sums:

    Suppose we have a geometric series with a constant rate r between terms. We let the sum SG represent the sum of the first n terms of that series such that,

    SG = k0 + k1 + . . . + kn-1 + kn

    Where k1 = k0(r), k2 = k1(r), and so on and so forth. Ultimately, using these definitions, it follows that,

    SG = k0 + k0(r) + k0(r)2 + . . . + k0(r)n-1

    If we multiply this sum by the constant rate r we find that,

    (r)SG = k0(r) + k0(r)2 + k0(r)3 + . . . + k0(r)n

    Finding the difference between the sums yields (the algebra is up to you),

    SG - (r)SG = k0 - k0(r)n

    Therefore,

    SG(1 - r) = k0(1 - rn)

    and consequently,

    SG = k0(1 - rn)/(1 - r)


    Hopefully you'll find this interesting or useful. If not, oh well.
     
  16. Jun 15, 2009 #15

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    1+ 3+ 5+ 7+ ...+ 93+ 95+ 97+ 99
    99+ 97+ 95+ 93+...+ 7+ 5+ 3+ 1

    What is the sum of each vertical column? How many columns are there? Notice that in adding each column and then getting the total you are adding "first 50 odd numbers" twice.

    Another way:
    1+ 3+ 5+ ...+ 45+ 47+ 49
    99+ 97+ 95+ ...+ 55+ 53+ 51

    What is the sum of each column? Now how many columns are there? So what is the sum of all the numbers. Notice that here, you have added each number exactly once.
     
  17. Jun 15, 2009 #16
    First, if you don't know how many numbers are there (the number n), find it using:
    [tex]a_n=a_1+(n-1)*d[/tex]

    where [itex]a_n[/itex] is the last number 49, [itex]a_1[/itex] is the first number 1, and [itex]d[/itex] is the difference between two numbers which equals 2.

    Just substitute and find n.

    Then, use the formula for finding the sum of n numbers of arithmetic progression:

    [tex]S_n=\frac{n}{2}*[a_1+a_n][/tex]

    Good luck.

    Regards.
     
  18. Jun 15, 2009 #17
    Thank you, this helps a lot.
     
  19. Jun 15, 2009 #18

    Thanks, I think I got it now.
     
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