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Edexcel FP1 Thread - 20th May, 2016

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swear the F1 papers are harder than the IAL FP1 papers??? had a 4+ mark difference between the 2 papers for jan 14
Original post by tazza ma razza
swear the F1 papers are harder than the IAL FP1 papers??? had a 4+ mark difference between the 2 papers for jan 14


Careful now, some of those papers have content that isn't in your FP1 spec (just one teeny bit of content, but that might be what caused the mark drop). However, F1/FP1 papers are not inherently harder than one another, one may be harder iin a particular year as a coincidence, but it is not intended.
Original post by Zacken
Careful now, some of those papers have content that isn't in your FP1 spec (just one teeny bit of content, but that might be what caused the mark drop). However, F1/FP1 papers are not inherently harder than one another, one may be harder iin a particular year as a coincidence, but it is not intended.


Nah IAL papers are defo harder tbf
Original post by BBeyond
Nah IAL papers are defo harder tbf


yh the jan or june 14 for s2 was the only one where it was p1ss easy incomparison to the bog standard.

Tho the F1 papers seem harder then ial which are harder than the r papers which are harder than the normal papers imo

annoyingly boundaries remain the same so a hard paper means you still need a high mark for the 90+ ums (hoping this isn;t the case in 2 weeks lol)
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Could someone explain the last 2 terms for me please?
Original post by Chirstos Ioannou
Screenshot_1.png
Could someone explain the last 2 terms for me please?


∑r=0n(2n+1)=(2n+1)∑r=0n1=(2n+1)(1+1+1+⋯+1⏟(n+1) times)=(2n+1)(n+1)\displaystyle \sum_{r=0}^{n} (2n+1) = (2n+1)\sum_{r=0}^{n} 1 = (2n+1)(\underbrace{1 + 1 + 1 + \cdots + 1}_{(n+1) \, \text{times}}) = (2n+1)(n+1)

There are (n+1) 1's because the sum starts from 0.
Original post by Zacken
∑r=0n(2n+1)=(2n+1)∑r=0n1=(2n+1)(1+1+1+⋯+1⏟(n+1) times)=(2n+1)(n+1)\displaystyle \sum_{r=0}^{n} (2n+1) = (2n+1)\sum_{r=0}^{n} 1 = (2n+1)(\underbrace{1 + 1 + 1 + \cdots + 1}_{(n+1) \, \text{times}}) = (2n+1)(n+1)

There are (n+1) 1's because the sum starts from 0.


So the reason you are able to pull (2n+1) out of the sum expression is because you can think of it as a number and not a variable?
Original post by Chirstos Ioannou
So the reason you are able to pull (2n+1) out of the sum expression is because you can think of it as a number and not a variable?


It's not a variable. You can't pull 'r' out because you're summing over r.

Basically, in a sum ∑r=0n\sum_{r=0}^{n} you can pull out anything except terms that contain the letter that is at the bottom of the sum, so in this case, you can't pull out r, but you can pull out anything else.

If you had ∑k=0n\sum_{k=0}^{n} you can pull out anything except k, etc...
Original post by Zacken
It's not a variable. You can't pull 'r' out because you're summing over r.

Basically, in a sum ∑r=0n\sum_{r=0}^{n} you can pull out anything except terms that contain the letter that is at the bottom of the sum, so in this case, you can't pull out r, but you can pull out anything else.

If you had ∑k=0n\sum_{k=0}^{n} you can pull out anything except k, etc...


Yeap got it, thanks :wink:
Original post by Chirstos Ioannou
Yeap got it, thanks :wink:


No worries.
Just came over to this thread, I did the exam last year so ask away!
But why are the first two terms the same as in the formula since the sum starts from r=0?
Original post by Zacken
No worries.


But why are the first two terms the same as in the formula since the sum starts from r=0?
Original post by Chirstos Ioannou
But why are the first two terms the same as in the formula since the sum starts from r=0?


Because you can split them up like so:

∑r=0n(r2−2r)=(02−2(0))+∑r=1n(r2−2r)=0+∑r=1n(r2−2r)\displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).
Original post by Zacken
Because you can split them up like so:

∑r=0n(r2−2r)=(02−2(0))+∑r=1n(r2−2r)=0+∑r=1n(r2−2r)\displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).


Exactly
Original post by Zacken
Because you can split them up like so:

∑r=0n(r2−2r)=(02−2(0))+∑r=1n(r2−2r)=0+∑r=1n(r2−2r)\displaystyle \sum_{r=0}^{n} (r^2 - 2r) = (0^2 - 2(0)) + \sum_{r=1}^n (r^2 - 2r) = 0 + \sum_{r=1}^{n} (r^2 - 2r).

i.e: the zeroth term is 0 for the first two terms, so summing from 0 or summing from 1 is the same thing (in this case).


So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?
Original post by Redcoats
Exactly


?

Original post by Chirstos Ioannou
So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?


Yeah.
Original post by Zacken
?



Yeah.


Ok thank you :wink:

He just agreed to your explanation.
Original post by Chirstos Ioannou
So what if it had a +5 in the sum then you would get the same result but with a +5 as the first term?


Yes - it is really useful to write out the sequence; I did that all over my FP1 exam!
Original post by Chirstos Ioannou
Ok thank you :wink:

He just agreed to your explanation.


Yeah, if you find yourself getting confused by the sigma notation, write out the series explicitly.

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