Summing Series - OCR FP1

Hi guys,
Sorry to ask some rather basic questions regarding this but I've just started Further Maths this year and it doesn't look rosy so far.

We haven't done C2 yet (apperently it does contain a section on sequences/series) so a little bit in to the deep end with this topic with regards to my maths ability.

I understand the Sigma notation, and how to sum an arithmetic series so far if it's not too hard but anything harder than that and I'm starting to struggle.

For those with access to the Quadling FP1 book I've taken some problems from Exercise 2A to show the kinds of questions I'm finding tricky.

$\displaystyle\sum_{i=1}^n (2r+1)$

It's stupidly easy I know, but once I break it down (law of addition/multiplication according to the teacher - please correct if I'm completely wrong) to:

$\displaystyle\sum_{i=1}^n (2r) \displaystyle\sum_{i=1}^n (1)$

and the multitude of brackets come in, I'm completely lost (have I even got the right approach in the first place?!).

Since this is presumably on the easier side of FP1 I'm thinking that I'm just not good enough to do FM and I should bail out before it's too late.

Thanks for all your help guys.

Hi guys,
Sorry to ask some rather basic questions regarding this but I've just started Further Maths this year and it doesn't look rosy so far.

We haven't done C2 yet (apperently it does contain a section on sequences/series) so a little bit in to the deep end with this topic with regards to my maths ability.

I understand the Sigma notation, and how to sum an arithmetic series so far if it's not too hard but anything harder than that and I'm starting to struggle.

For those with access to the Quadling FP1 book I've taken some problems from Exercise 2A to show the kinds of questions I'm finding tricky.

$\displaystyle\sum_{i=1}^n (2r+1)$

It's stupidly easy I know, but once I break it down (law of addition/multiplication according to the teacher - please correct if I'm completely wrong) to:

$\displaystyle\sum_{i=1}^n (2r) \displaystyle\sum_{i=1}^n (1)$

and the multitude of brackets come in, I'm completely lost (have I even got the right approach in the first place?!).

Since this is presumably on the easier side of FP1 I'm thinking that I'm just not good enough to do FM and I should bail out before it's too late.

Thanks for all your help guys.

$\displaystyle\sum_{i=1}^n (2r+1)=\displaystyle\sum_{i=1}^n (2r) +\displaystyle\sum_{i=1}^n (1)=2\displaystyle\sum_{i=1}^n (r) +\displaystyle\sum_{i=1}^n (1)$
$\displaystyle\sum_{i=1}^n (1)$ is just n and
$\displaystyle\sum_{i=1}^n (r)$ is a standard result which you need to learn. Work it out as the sum of an arithmetic series with first term 1 and common difference 1 if you do not know it.

Thanks guys.



Right.

So:

$\displaystyle\sum_{i=1}^n (r) = \frac{1}{2} n(n+1)$

Since there are 2:
$\frac{1}{2} n(n+1) + \frac{1}{2} n(n+1)$

Giving:
$n(n+1)$

Being the idiot that I am, I've ended up with the above answer instead of $n(n+2)$

Maybe I should give up FM and just stick to normal maths!

You still have an 'n' left over from the sum of (1) which you need to add on

Thanks. So 1 = n therefore it's $n(n+2)$

Sorry to be a real pain but I just want to check if my methodology is along the right lines or not.

Question 2C now:
$\displaystyle\sum_{r=1}^n = (3r^2+1)$

Goes to:

$3\displaystyle\sum_{r=1}^n = r^2 + \displaystyle\sum_{r=1}^n = n$

What's the easiest way of dealing with $3\displaystyle\sum_{r=1}^n = r^2$ without having to go through the process of expanding etc?

I'm probably missing an important step/technique here.

Sorry for all the questions guys.

Thanks. So 1 = n therefore it's $n(n+2)$

Sorry to be a real pain but I just want to check if my methodology is along the right lines or not.

Question 2C now:
$\displaystyle\sum_{r=1}^n = (3r^2+1)$

Goes to:

$3\displaystyle\sum_{r=1} ^ n r^2 + \displaystyle\sum_{r=1}^n 1$

What's the easiest way of dealing with $3\displaystyle\sum_{r=1}^n = r^2$ without having to go through the process of expanding etc?

I'm probably missing an important step/technique here.

Sorry for all the questions guys.

Thanks. So 1 = n therefore it's $n(n+2)$

Sorry to be a real pain but I just want to check if my methodology is along the right lines or not.

Question 2C now:
$\displaystyle\sum_{r=1}^n = (3r^2+1)$