A level further maths help needed!

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 $\displaystyle \sum_{r=1}^{3n} (r^2-r+3)$ ?

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 ... ???

I have no idea what the upper limit is

And would you put (2n+1) into the equation?

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

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.