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Thread: Find the sum, arithmetic sequence

  1. #1
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    Find the sum, arithmetic sequence

    determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.

    this is how I thought:

    2+4+6+8+...+1000

    an=a1+(n-1)*d

    1000=2+(n-1)*2
    1000=2+2n-2
    1000-2+2=2n
    n=500

    sum=(n/2)*(a1+an)=(500/2)*(2+1000)
    sum=250500

    Then I tought

    1+3+5+7+...+1000

    an=a1+(n-1)*d
    1000=1+(n-1)*2
    1000=1+2n-2
    1000=-1+2n
    1001=2n
    n=500,5

    sum=(500,5/2)*(1+1000)
    sum=250500,25

    250500,25-
    250500=0,25
    The answer is 1000. How did I Count wrong??




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  2. #2
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    Re: Find the sum, arithmetic sequence

    Quote Originally Posted by selin View Post
    determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.
    this is how I thought:

    2+4+6+8+...+1000
    an=a1+(n-1)*d
    1000=2+(n-1)*2
    1000=2+2n-2
    1000-2+2=2n
    n=500

    sum=(n/2)*(a1+an)=(500/2)*(2+1000)
    sum=250500

    Then I tought

    1+3+5+7+...+1000

    an=a1+(n-1)*d
    1000=1+(n-1)*2
    1000=1+2n-2
    1000=-1+2n
    1001=2n
    n=500,5

    sum=(500,5/2)*(1+1000)
    sum=250500,25

    250500,25-
    250500=0,25
    The answer is 1000. How did I Count wrong??
    $\sum\limits_{k = 1}^{{{10}^3}} {(2k)} = 2\sum\limits_{k = 1}^{{{10}^3}} {(k)} = ?$

    $\sum\limits_{k = 1}^{{{10}^3}} {(2k - 1)} = 2\sum\limits_{k = 1}^{{{10}^3}} {(k)} - \sum\limits_{k = 1}^{{{10}^3}} {(1)} = ?$
    Thanks from topsquark
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  3. #3
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    Re: Find the sum, arithmetic sequence

    Quote Originally Posted by selin View Post
    determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.

    this is how I thought:

    2+4+6+8+...+1000 No. This is only the first 500 even numbers.

    an=a1+(n-1)*d

    1000=2+(n-1)*2
    1000=2+2n-2
    1000-2+2=2n
    n=500 No, you are told n=1000

    sum=(n/2)*(a1+an)=(500/2)*(2+1000)
    Use sum = (n/2)(a1 + (n-1)d) where n=1000, a=2 and d=2
    sum=250500

    Then I tought

    1+3+5+7+...+1000
    No. 1000 is not odd.

    an=a1+(n-1)*d
    1000=1+(n-1)*2
    1000=1+2n-2
    1000=-1+2n
    1001=2n
    n=500,5 Firstly n must be an integer. Secondly you are told that n=1000

    sum=(500,5/2)*(1+1000)
    sum=250500,25 How could this not be an integer?? Use sum = (n/2)(a1+(n-1)d) where a=1, n=1000, d=2

    250500,25-
    250500=0,25
    The answer is 1000. How did I Count wrong??




    See comments in red above.
    If you make the changes I suggested you will get the correct answer. But that's not the most efficient way of finding the solution to the problem.

    Think of it this way:
    Sum of first 1000 odd numbers = 1 + 3 + 5 + 7 + . until you have 1000 terms
    Sum of first 1000 even numbers = 2 + 4 + 6 + 8 + until you have 1000 terms = (1+1)+(3+1)+(5+1)+(7+1) + until you have 1000 pairs

    Notice that each term in the "even" sum is 1 more than the corresponding term in the "odd" sum.

    So the "even" sum will be 1000 x 1 = 1000 more than the "odd" sum.
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  4. #4
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    Re: Find the sum, arithmetic sequence

    Quote Originally Posted by selin View Post
    determine the difference between the sum of the 1000 first even positive integers and the sum of the first 1000 odd positive integers.
    The answer is 1000. How did I Count wrong??
    Do you know the so-called Gauss-Sum : $\sum\limits_{k = 1}^N k = \frac{{N(N + 1)}}{2}$ that is miss-named but that is another story.
    So $\sum\limits_{k = 1}^N {(2k)} = 2\sum\limits_{k = 1}^N {(k)} = 2\frac{{N(N + 1)}}{2} = N(N + 1)$

    If $\sum\limits_{k = 1}^N {(2k)} $ is the sum of the first $\displaystyle N$ even integers.
    Each number in that sum of even integers is $2k$.

    So look at the sum of the first $\displaystyle N $ odd integers $\displaystyle \sum\limits_{k = 1}^N {(2k -1) }
    = \sum\limits_{k = 1}^N {(2k) }-1 \sum\limits_{k = 1}^N {(1)} $

    Can you see that each odd number in that range is on less that the corresponding even number? There are $\displaystyle 10^3$ such numbers.
    SO?
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  5. #5
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    Re: Find the sum, arithmetic sequence

    My post #3 is the layman's version of what Plato just said.
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  6. #6
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    Re: Find the sum, arithmetic sequence

    I haven't learned what that E means. But I still managed to solve it, thank you anyway.
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  7. #7
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    Re: Find the sum, arithmetic sequence

    solved it! thanks!
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