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Math Help - Problem With A Sequence Proof

  1. #1
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    Problem With A Sequence Proof

    The theorem for finding the sum of n terms of an Arithmetic Sequence is:

    Sn = n/2[ 2a + (n - 1)d ] = n/2(a + an)


    However, I am having a problem with the proof of this which is given in the book. I will show the proof given and stop at the point where I'm having a problem understanding.

    Sn = a1 + a2 + a3 + ... + an ;...........................sum of first n terms

    = a + (a + d) + (a + 2d) + ... + [a + (n - 1)d] ;....uses an = a + (n - 1)d

    = (a + a + ... + a) + [d + 2d + ... + (n - 1)d] ;.....rearranging terms

    = na + d[1 + 2 + ... + (n - 1)] ;.........................there are n a terms and factors out d

    = na + d[ (n - 1)n / 2 ] ;..................................inside the brackets using property 1 + 2 + 3 + ... + n = n(n+1)/2


    It is at this point in the proof that I have a problem, since n(n-1)/2 is not the same as n(n+1)/2.
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  2. #2
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    Quote Originally Posted by spiritualfields View Post
    ...
    = na + d[1 + 2 + ... + (n - 1)] ;.........................there are n a terms and factors out d

    = na + d[ (n - 1)n / 2 ] ;..................................inside the brackets using property 1 + 2 + 3 + ... + n = n(n+1)/2

    It is at this point in the proof that I have a problem, since n(n-1)/2 is not the same as n(n+1)/2.
    Hello,

    you are right the sum of the first n summands is:

    s_n = n/2*(n-1)

    You don't have n summands but (n-1) summands. Use the formula to calculate the sum of the first (n-1) summands:

    s_(n-1) = (n-1)/2*((n-1) + 1) = (n-1)/2 * n = n(n-1)/2

    EB
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  3. #3
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    Earboth

    s_(n-1) = (n-1)/2*((n-1) + 1) = (n-1)/2 * n = n(n-1)/2
    Yes, now I can see how substituting (n - 1) for n causes the sign change. Thanks for the help.
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