Juliakinga
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How many terms are there in the series
3n sigma 1 (r^2-r+3)
- would it be 3?

B) write and simplify the (2n+1)th term and find the sum of the series
A) 1+2+3+...+3n

Would really appreciate any hell
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Plücker
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(Original post by Juliakinga)
How many terms are there in the series
3n sigma 1 (r^2-r+3)
- would it be 3?

B) write and simplify the (2n+1)th term and find the sum of the series
A) 1+2+3+...+3n

Would really appreciate any hell
Do you mean \displaystyle \sum_{r=1}^{3n} (r^2-r+3) ?
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Juliakinga
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(Original post by BuryMathsTutor)
Do you mean \displaystyle \sum_{r=1}^{3n} (r^2-r+3) ?
Yes thank you
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RDKGames
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(Original post by Juliakinga)
How many terms are there in the series
3n sigma 1 (r^2-r+3)
- would it be 3?

B) write and simplify the (2n+1)th term and find the sum of the series
A) 1+2+3+...+3n

Would really appreciate any hell
Not quite. If I pick n=2 then what's the upper limit? Hence how many terms are you adding in this sum?? So what's the number of terms in terms of n???

For (B), if the rth term is r^2-r+3, then the (2n+1)th term is ... ???
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Juliakinga
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(Original post by RDKGames)
Not quite. If I pick n=2 then what's the upper limit? Hence how many terms are you adding in this sum?? So what's the number of terms in terms of n???

For (B), if the rth term is r^2-r+3, then the (2n+1)th term is ... ???
I have no idea what the upper limit is
And would you put (2n+1) into the equation?
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RDKGames
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(Original post by Juliakinga)
I have no idea what the upper limit is
Come on now, the upper limit is 3n. If I pick n=2 then this just becomes 6. Since the lower limit is always 1, you will have exactly 6 terms.

So for a general upper limit 3n and lower limit 1, how many terms do we have ??
And would you put (2n+1) into the equation?
Simple as that.
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Juliakinga
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(Original post by RDKGames)
Come on now, the upper limit is 3n. If I pick n=2 then this just becomes 6. Since the lower limit is always 1, you will have exactly 6 terms.

So for a general upper limit 3n and lower limit 1, how many terms do we have ??


Simple as that.
So we would have up to 9 terms? I’m sorry I’m really bad at wordy things
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RDKGames
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(Original post by Juliakinga)
So we would have up to 9 terms? I’m sorry I’m really bad at wordy things
The sum \displaystyle \sum_{r=1}^{3n} will have exactly 3n terms.

I suggest you go back to the basics of how sigma notation works to ensure you understand why.
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Juliakinga
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(Original post by RDKGames)
The sum \displaystyle \sum_{r=1}^{3n} will have exactly 3n terms.

I suggest you go back to the basics of how sigma notation works to ensure you understand why.
Ohhh right
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