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Thread: Finding the sum of the series from n +1 to 49

  1. #1
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    Finding the sum of the series from n +1 to 49

    Question:
    The sum of the integers from 1 to n-1 is equal to the sum of integers from the n + 1 to 49. Find n




    I am able to find 1 to n - 1 which is
     s_{n-1}= \frac{1}{2}(n(n-1))


    but for (n+1) + (n+2) + ....... + 49


    I try this:  (n + 1) + (n+2) + (n+3) + ....... + 49
    a = n+1 common difference = 1
    nth term = 2n
    thus
    49th term   = 2(49) = 98


    let n = 1
    first term = 2(1) = 2


    thus sum of n+1 to 49


     s_{49}= \frac{49}{2}(2(2)+(49-1)(1)) = 49(26)=1274
    thus
      \frac{1}{2}(n(n-1)) = 1274




    thus n = -49.9801....... or n = 50.980...
    but n is an integer
    please can some point me in the right direction
    how do I found the first term, common difference and number of terms for the series from (n+1) to 49




    Thank you in advance
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  2. #2
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    Re: Finding the sum of the series from n +1 to 49

    what on earth.... :O

    $\displaystyle \sum_1^{n-1} ~k = \dfrac{n(n-1)}{2}$

    $\displaystyle \sum_1^{49} ~k = \dfrac{(49)(50)}{2} = 1225$

    $\displaystyle \sum_1^{n} ~k = \dfrac{n(n+1)}{2}$

    $1225 = \dfrac{n(n+1)}{2} + \displaystyle \sum_{n+1}^{49} ~k$

    if $\displaystyle \sum_1^{n-1} ~k = \displaystyle \sum_{n+1}^{49} ~k$


    $1225 = \dfrac{n(n+1)}{2} + \dfrac{n(n-1)}{2} = \dfrac{n}{2}(n+1+n-1) = n^2$

    $n = \sqrt{1225} = 35$
    Thanks from topsquark
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    Re: Finding the sum of the series from n +1 to 49

    [1+2+3+....+(n-1)] n [(n+1)+(n+2)+....+49]

    (n-1)n/2 = 49(50)/2 - n(n+1)/2 : solve for n

    aNUTter way to look at Romsek's...
    Last edited by DenisB; Apr 19th 2017 at 07:47 PM.
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  4. #4
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    Re: Finding the sum of the series from n +1 to 49

    Quote Originally Posted by bigmansouf View Post
    Question:
    The sum of the integers from 1 to n-1 is equal to the sum of integers from the n + 1 to 49. Find n
    $\sum\limits_{k = 1}^{49} k = \frac{{49 \cdot 50}}{2} = 49 \cdot 25$
    If $1<J<49,$ then $\boxed{\sum\limits_{k = 1}^{J-1} k }+J+\boxed{\sum\limits_{k =J+ 1}^{49} k }$.
    We are given that the two boxed sums are equals thus that gives:
    $2\dfrac{(J-1)(J)}{2}+J=49 \cdot 25$ This gives the solution $N=35$
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