sum of the series

[xc630]

Hi I am having toruble with this sereis. I need to find the sum of the integers except for multiples of 3.

The series is 1+2,+4,+5,+7,+8,+10+11,+...+299

The correct answer should be 30000 but I got a different answer. I found the sum of 1-299 to be 44850. Then I found the amount of terms that are multiples of 3 which tunred out to be 33. Usign this I foudn the sum for the 33 terms that were multiples of 3 and I got 4950. I then tried subtracting the two sums but I got it wrong. Am I got the right track?

 

[skeeter]

1 + 2 + 3 + ... + n = n(n+1)/2


1 + 2 + 3 + ... + 297 + 298 + 299 = (299)(300)/2

3 + 6 + 9 + ... + 294 + 297 = 3(1 + 2 + 3 + ... + 98 + 99) = 3(99)(100)/2

can you finish?

 

[Gene]

Good thinking, bad counting. The count of 33 is wrong. try that again.
Also the sum of 33 multiples of 3 would be
3*(33*34/2) = 1683

 

[galactus]

\(\displaystyle \L\\\sum_{n=1}^{299}(n)-\sum_{n=1}^{99}(3n)\)

=\(\displaystyle \L\\\frac{n(n+1)}{2}-3(\frac{n(n+1)}{2})\)

=\(\displaystyle \L\\\frac{(299)(300)}{2}-3(\frac{99(100)}{2})\)

 

[xc630]

Oh ok, n should be 33. I got it. I think I had the sum of the sries formula right though too becuase it's arithmetic. Thanks.

 

[Gene]

I'm don't know what you are saying about n=33. There are 99 multiples of 3 between 1 and 299 to be subtracted, not 33.
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Gene