FP1 Question

I've just done section B of last year's FP1 paper as part of my mock and there's a question I couldn't figure out.

The question is on summation of series, which we haven't actually covered yet, but I get another shot at the question of Friday and I kind of just want to show that I can do it anyway.
The question (more or less - I memorised it as well as I could):

45/(39*41*43) + 46/(40*42*44) + 47/(41*43*45) + ... + 105/(99*101*103)

What I worked out so far:
A general formula of n/((n-6)(n-4)(n-2))

I had a friend try and explain it to me and he did a fairly good job of it but, since I haven't learnt about this yet, it didn't make a whole lot of sense.

Could someone just walk me through this? Thanks!

I've just done section B of last year's FP1 paper as part of my mock and there's a question I couldn't figure out.

The question is on summation of series, which we haven't actually covered yet, but I get another shot at the question of Friday and I kind of just want to show that I can do it anyway.
The question (more or less - I memorised it as well as I could):

45/(39*41*43) + 46/(40*42*44) + 47/(41*43*45) + ... + 105/(99*101*103)

What I worked out so far:
A general formula of n/((n-6)(n-4)(n-2))

I had a friend try and explain it to me and he did a fairly good job of it but, since I haven't learnt about this yet, it didn't make a whole lot of sense.

Could someone just walk me through this? Thanks!

Look at the "Method of Differences" for summation.

Please give slightly more info. Did it ask for you to evaluate the summation? If so it will have given you some general formula to apply.

And yes method of differences is what you want.

Look at the "Method of Differences" for summation.

Yes, I need to sum the values. There was no formula at all, only the series, which is why I have no idea how to do it. I've never seen method of differences without a given formula, and the one I made up for it doesn't work because it's only one fraction.

I think you copied the question wrongly... The general expression for this sum is unusual to be in any FP1 paper, but I could be wrong.

https://www.wolframalpha.com/input/?i=sum+n%2F((n-6)(n-4)(n-2))+from+n%3D45

The formula in terms of n is just my own thing that I worked out. It wasn't part of the question. All that was in the question was the series and the instruction to find its sum. No formula or anything.
I'm almost certain that the question is copied correctly. The pattern was definitely this. The only thing I'm slightly unsure about are the bounds.

Thanks!

Right, so in the actual paper. The question was split into several parts. The first part was to express

$\dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}=\dfrac{3}{4(2r-1)}-\dfrac{1}{2r+1}+\dfrac{1}{4(2r+3 )}$ as partial fractions.

Second part was to find $\displaystyle \sum_{r=1}^n\dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}$ using the method of differences utilising the partial fraction expansion.

Third part was to find the limit as $n \to \infty$ of the sum.

And fourth part was to compute$\displaystyle \sum_{20}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}$ by writing it as $\displaystyle \sum_{r=1}^{50} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)} - \sum_{r=1}^{19} \dfrac{2r+5}{(2r-1)(2r+1)(2r+3)}$

and using evaluating those two by using the previous expression for $n=50$ and $n=19$.

Now that it's in this more structured form, try having a go? Although you should look up "method of differences" before. The final answer should be ~0.00813 or something along those lines.

Hmm interesting. I'll give it a go

Is this old spec? Never come across this is fp1. Although i know it from c4.

OCR MEI 2016 FP1 paper.

Ah MEI. That'd explain it.
Kinda wish we were doing MEI now. Sounds really fun.

Hmm interesting. I'll give it a go

Is this old spec? Never come across this is fp1. Although i know it from c4.