15 Jun '07 15:32>
thats right and we will leave it at that because it is far to complicated to explain otherwise.
Originally posted by sonhouseThere 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.
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
Originally posted by Drew LYeah 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.
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
Originally posted by kbaumenWow, what did those kids have to do if they started acting up or something? 🙂
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.