1. Joined
    01 May '07
    Moves
    27311
    15 Jun '07 15:32
    thats right and we will leave it at that because it is far to complicated to explain otherwise.
  2. Sigulda, Latvia
    Joined
    30 Aug '06
    Moves
    4048
    15 Jun '07 16:18
    Originally posted by sonhouse
    I got 1050. I did it in groups of 5, 18+20+22+24+26 and found each group starting with 110 sum going up by 50 so it was easy to truncate the series, 110+160+210+260+310= 1050
    There is an easier way to do this. Just sum the first integer with the last. You get 12 pairs with equal sums, count them all together and add the one integer that was left as the number of integers given is odd.

    For example 1+2+3+...+99+100+101=50*((1+101)+(2+100)+...+(50+52))+51=5151
  3. Joined
    01 May '07
    Moves
    27311
    15 Jun '07 18:49
    the way we had to do it on my final was as follows :

    An = A1 + (n-1)d

    where
    An = any term
    A1 = first term
    n = number of terms in the sequence
    d = arithmatic difference between terms

    so

    An = 18 + (25-1)2
    An = 18 + 48
    An = 66 < last term in the sequence

    then find the sum using the sum formula for any arithmatic sequence

    Sn = (n/2) (A1 + An)

    where
    Sn = sum
    n = number of terms in sequence
    A1 = first term
    An = last term

    so

    Sn = (25/2) (18 + 66)
    Sn = 25 ( 9 + 33)
    Sn = 25 ( 42 )
    Sn = 1050
  4. Sigulda, Latvia
    Joined
    30 Aug '06
    Moves
    4048
    15 Jun '07 19:321 edit
    Originally posted by Drew L
    the way we had to do it on my final was as follows :

    An = A1 + (n-1)d

    where
    An = any term
    A1 = first term
    n = number of terms in the sequence
    d = arithmatic difference between terms

    so

    An = 18 + (25-1)2
    An = 18 + 48
    An = 66 < last term in the sequence

    then find the sum using the sum formula for any arithmatic sequence

    Sn = (n/2) ( ...[text shortened]... rm
    An = last term

    so

    Sn = (25/2) (18 + 66)
    Sn = 25 ( 9 + 33)
    Sn = 25 ( 42 )
    Sn = 1050
    Yeah that is another way to do so. But the way I described was used by Karl Gauss when he was a kid. For behaving, he was asked by his teacher to sum all natural numbers from 1 to 100. Surprisingly, he had the answer in a matter of seconds, and the way I described it is how he did it. Of course they way you describe is taught in school and I've learned it as well and I think both ways are good enough.
  5. Joined
    31 May '06
    Moves
    867
    16 Jun '07 03:43
    Originally posted by kbaumen
    Yeah that is another way to do so. But the way I described was used by Karl Gauss when he was a kid. For behaving, he was asked by his teacher to sum all natural numbers from 1 to 100. Surprisingly, he had the answer in a matter of seconds, and the way I described it is how he did it. Of course they way you describe is taught in school and I've learned it as well and I think both ways are good enough.
    Wow, what did those kids have to do if they started acting up or something? 🙂
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