Edexcel FP1 Thread - 20th May, 2016

How would you do part d?

$\sum\limits_{r=0}^n (2n+1)$ = $\sum\limits_{r=1}^n (2n+1)$ + 2(0) +1

I believe

How did you get to that answer?

Sorry my mistake - not sure what do you since you aren't summing "r" terms

do you think it might be n(2n+1) + 1 ?

I thought it would be n(2n+1).

Hi, I am stuck on question 10(ii) of the June 2013 R paper. I understand that summation of the $r^2$ term and the $-2r$ term, but don't understand what the summation of $\sum\limits_{r=0}^n (2n+1)$ will produce. I initially thought it will be equal to $n(2n+1)$ but I am pretty sure that's incorrect. Does the summation from 0 matter or influence the result?

when n=0, 2n+1 = 1 though?
then from n=1 to n=n it's a sum of n times 2n+1

hence n(2n+1) + 1

Not sure though

when n=0, 2n+1 = 1 though?
then from n=1 to n=n it's a sum of n times 2n+1

hence n(2n+1) + 1

Not sure though

See my post above. Remember that you are summing over r.

So when r=0 we have (2n+1) = (2n+1) in no way are we ever setting n=0.

I'll give you two ways to look at it:

1. $\displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)$

2. $\displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}$.

Since in this case you have $\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1$

Then you have $f(r) = 1$ being added together $n+1$ times to get a total of $(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1)$.

ah ok,it's just strange that you're summing for r, but the sum is of "n"s... I guess you just remember to do n * the thing in the sum, and add 1 * the thing in the sum (for when r=0) ?

Thanks so much, this really helped! I understand the question using the second way, but the first way is a bit confusing. For the 0th term could you write it as $\displaystyle \sum_{r=0}^0 (2n+1) = (2n+1) \sum_{r=0}^0 1$ and then isn't $\displaystyle \sum_{r=0}^0 1 = 0$, so the whole term is 0?

No, that doesn't make sense, the first term has nothing to do with a sum.

$\displaystyle \sum_{r=0}^n f(r) = f(0) + f(1) + \cdots$

Here, we have $f(0) = 2n+1$, that's the 0th term. There's no sum coefficient or anything.

Has anyone completed the crashmaths FP1 past papers? I feel like there are multiple mistakes in the first one.

Has anyone completed the crashmaths FP1 past papers? I feel like there are multiple mistakes in the first one.

Having completed the paper I, myself, noticed one mistake in the mark scheme of the 'Set A' paper. I did not notice any other mistakes in it apart from one. I think the paper was very good - you can't expect everything to be flawless ultimately, CrashMaths/Aaran has other things to worry about than one FP1 paper, all he's trying to do is make a paper which is giving us an extra opportunity to see where we are at with our knowledge and understanding of the FP1 module. I'd say you should be slightly more grateful.

If you don't mind me asking, what mistakes did you notice? I only noticed one arithmetic error in the mark scheme which I pointed out and was amended quickly, otherwise apart from that I did not find any others, and found the paper actually very useful..

where do you think the mistakes are?