C1 sequences and series Watch

kazaa
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This is a very stupid question, but when you use the sigma notation for working out the sum of something. eg. r=20 on top and r=5 on bottom and then something like (2r+4), what do you use for n? When r=1 on the bottom, n is the r on top, 20...but when r is 5??
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Trevor 12345
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Its not a stupid question.


you would find the answer to your question ::r=20 on top and r=5 on bottom and (2r+4).

by working out

:r=20 on top and r=1 on bottom with (2r+4) and r=4 on top and r=1 on bottom with (2r+4); then subtract the Sums...

mrm.
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kazaa
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oh...i didnt know about that...why r=4 on top?
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Dez
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As far as I'm aware, the C1 sum formulae only works from 1 to n, so if you had to go from five, I'd fudge it to something like this:

\displaystyle\huge\sum_{r=1}^{20  } (2r+4) - \sum_{r=1}^4 (2r+4)

That'll get rid of the first four terms, leaving you with 5 <= n <= 20.
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kazaa
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For this particular example, they give r=30 on top and r=10 on bottom, with (7+2r). So a=27, d=2 and n=30-9=21...where did they get the 21 from? :confused: Am i going insane and not realising something really obvious?
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Trevor 12345
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(Original post by kazaa)
oh...i didnt know about that...why r=4 on top?

so that you are left with the sum of terms 5 upto 20
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Trevor 12345
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(Original post by kazaa)
For this particular example, they give r=30 on top and r=10 on bottom, with (7+2r). So a=27, d=2 and n=30-9=21...where did they get the 21 from? :confused: Am i going insane and not realising something really obvious?

ok there are 2 ways of doing these things either the sums as suggested above or indeed just using the sum of an arithmetic series.

From r=10 to r=30 means from (and including) the 10th to the 30th term. That makes 21 terms in total in general fro r=a to r=b will have b-a+1 terms in total
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kazaa
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Oooh I see! I completely forgot about that! Thanks!!
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