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FP1 edexcel specification query

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Original post by Jai Sandhu
Yes, it gives you terms in n as n is a constant, are you clear as to why we have n+1 though?


Err....Maybe...not sure!
Original post by Mutleybm1996
Err....Maybe...not sure!


bro what paper is q1 from? pls :tongue:
Original post by BBeyond
bro what paper is q1 from? pls :tongue:


January 2014 F1 Paper.


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Original post by Jai Sandhu
ImageUploadedByStudent Room1431384933.076263.jpg

You can be the judge of that!


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I approve of all except your Z\mathbb{Z} and your lambda, the first has far too skinny a body and well your lambda... :wink:
Original post by Mutleybm1996
Err....Maybe...not sure!


n+1 is constant, it doesn't "depend" on r
Original post by Mutleybm1996
Err....Maybe...not sure!


Alright, the standard forms givin to you for summing series are written in this form.

∑r=1n1=n\sum_{r=1}^n 1 = n

However, we are summing from r=0, this means that there is an additional 1 which we need to add on resulting in.

∑r=0n1=n+1\sum_{r=0}^n 1 = n + 1

Remember, n is not something we are summing it is merely a result from a sum, therefore we can treat it as a constant in the same way that.

∑r=1n2r\sum_{r=1}^n 2r

is the the same as

2×∑r=1nr2\times\sum_{r=1}^n r

In this case we have:

∑r=0n2n+1\sum_{r=0}^n 2n+1

which is the same as

2n+1×∑r=0n12n+1\times\sum_{r=0}^n 1
Original post by lizard54142
I approve of all except your Z\mathbb{Z} and your lambda, the first has far too skinny a body and well your lambda... :wink:


My Z kinda messed up, it should look like the one in the bottom right.
Original post by kingaaran
January 2014 F1 Paper.


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cheers :biggrin:
Original post by Jai Sandhu
My Z kinda messed up, it should look like the one in the bottom right.


Ah yes, I missed that one.
Original post by Jai Sandhu
Alright, the standard forms givin to you for summing series are written in this form.

∑r=1n1=n\sum_{r=1}^n 1 = n

However, we are summing from r=0, this means that there is an additional 1 which we need to add on resulting in.

∑r=0n1=n+1\sum_{r=0}^n 1 = n + 1

Remember, n is not something we are summing it is merely a result from a sum, therefore we can treat it as a constant in the same way that.

∑r=1n2r\sum_{r=1}^n 2r

is the the same as

2×∑r=1nr2\times\sum_{r=1}^n r

In this case we have:

∑r=0n2n+1\sum_{r=0}^n 2n+1

which is the same as

2n+1×∑r=0n12n+1\times\sum_{r=0}^n 1


Original post by lizard54142
n+1 is constant, it doesn't "depend" on r


Thank you all,
is it possible for you to quickly check through?
I'm pretty sure i understand it now,
just to check though...

So, anything with (Xn+1) in the formula and I'd do (Xn+1) times sum(r=0 to n) of 1.
then add (2n+1)(n+1) to the end of the standard results and follow this through?

Alas, that phrasing is not very mathematical I'm afraid, but does that make sense?
1.PNG39.4KB
2.PNG23.8KB
Original post by Mutleybm1996
Thank you all,
is it possible for you to quickly check through?
I'm pretty sure i understand it now,
just to check though...

So, anything with (Xn+1) in the formula and I'd do (Xn+1) times sum(r=0 to n) of 1.
then add (2n+1)(n+1) to the end of the standard results and follow this through?

Alas, that phrasing is not very mathematical I'm afraid, but does that make sense?


Sounds reasonable. But that is only if it is the sum from r=0 to n. ANY terms which contain an "n" such as n^2, n^3, even e^n (I doubt the last one would ever appear haha) are all constant, which only depend on what the limit of the summation is, not the "summand" (i.e. the r's).
Original post by lizard54142
Sounds reasonable. But that is only if it is the sum from r=0 to n. ANY terms which contain an "n" such as n^2, n^3, even e^n (I doubt the last one would ever appear haha) are all constant, which only depend on what the limit of the summation is, not the "summand" (i.e. the r's).


Would it be possible to ask for a further example?


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Original post by Mutleybm1996
Would it be possible to ask for a further example?


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I'll type up a more mathematical example for you tomorrow, for your notes. I'm too tired now I need sleep :frown:
Original post by lizard54142
I'll type up a more mathematical example for you tomorrow, for your notes. I'm too tired now I need sleep :frown:


This


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Original post by Jai Sandhu


I have M2 tomorrow, yet I still stay up stupidly late :s-smilie:
Original post by Mutleybm1996
Would it be possible to ask for a further example?


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∑r=1n1=n\displaystyle\sum_{r=1}^n 1 = n

∑r=011=1\displaystyle\sum_{r=0}^1 1 = 1

⇒∑r=0n1=n+1\Rightarrow \displaystyle\sum_{r=0}^n 1 = n + 1

Let f(n)f(n) be a function of nn

∑r=0nf(n)=f(n)∑r=0n1=f(n).(n+1)\displaystyle\sum_{r=0}^n f(n) = f(n) \displaystyle\sum_{r=0}^n 1 = f(n).(n+1)
Original post by lizard54142
I have M2 tomorrow, yet I still stay up stupidly late :s-smilie:


I also have M2 (OCR) tomorrow.
Original post by Jai Sandhu
I also have M2 (OCR) tomorrow.


How confident are you? I am on MEI.
Original post by lizard54142
How confident are you? I am on MEI.


I am non-MEI, ive done every past paper atleast twice, if I dont get 100 UMS ill be gutted.
Original post by Jai Sandhu
I am non-MEI, ive done every past paper atleast twice, if I dont get 100 UMS ill be gutted.


Same, aiming for that 100, although M2 is the one module you can make so many silly mistakes on.

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