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FP1 hard question

Hello, I know this question has been answered several times but I just cannot seem to understand the mark scheme or any of the explanations? It is the second part of course - I have no idea how the 2n is manipulated or how it is turned into what it is turned into. If someone could try and explain it I'd be grateful!! Thank you :smile:
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Reply 1
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Zacken
22
Original post by iMacJack
Hello, I know this question has been answered several times but I just cannot seem to understand the mark scheme or any of the explanations? It is the second part of course - I have no idea how the 2n is manipulated or how it is turned into what it is turned into. If someone could try and explain it I'd be grateful!! Thank you :smile:


The (2n+1)(2n+1) is just a constant. So ∑r=0n(2n+1)=(2n+1)∑r=0n(1)=(2n+1)(n+1)\sum_{r=0}^{n} (2n+1) = (2n+1) \sum_{r=0}^n (1) = (2n+1)(n+1)
Reply 2
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iMacJack
OP
16
Original post by Zacken
The (2n+1)(2n+1) is just a constant. So ∑r=0n(2n+1)=(2n+1)∑r=0n(1)=(2n+1)(n+1)\sum_{r=0}^{n} (2n+1) = (2n+1) \sum_{r=0}^n (1) = (2n+1)(n+1)


Am I right in assuming if the series was from r = 1 to n instead of r = 0 to n it would be (2n+1)(n) instead of (2n+1)(n+1) basically, is it (n+1) because the extra term is involved due to it beginning from 0 instead of one? Thank you :smile:
Reply 3
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Zacken
22
Original post by iMacJack
Am I right in assuming if the series was from r = 1 to n instead of r = 0 to n it would be (2n+1)(n) instead of (2n+1)(n+1) basically, is it (n+1) because the extra term is involved due to it beginning from 0 instead of one? Thank you :smile:


Yep! Bang on. :biggrin:

I've answered this very question at least four times this past month, seems like it trips up a lot of people. :lol:
Reply 4
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iMacJack
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Original post by Zacken
Yep! Bang on. :biggrin:

I've answered this very question at least four times this past month, seems like it trips up a lot of people. :lol:

Haha yes it does indeed!! What if I just dealt with the 2n separately to the + 1 instead of combining it to 2n+1(n+1)?
Reply 5
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Zacken
22
Original post by iMacJack
Haha yes it does indeed!! What if I just dealt with the 2n separately to the + 1 instead of combining it to 2n+1(n+1)?


Well, you'd get the same thing:

Unparseable latex formula:

\displaystyle[br]\begin{align*}\sum_{r=0}^n (2n) + \sum_{r=0}^n (1) &= 2n \sum_{r=0}^n (1) + \sum_{r=0}^n (1) \\&= (2n)(n+1) + (1)(n+1)\\& = (2n+1)(n+1)\end{align*}

Reply 6
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iMacJack
OP
16
Original post by Zacken
Well, you'd get the same thing:

∑r=0n(2n)+∑r=0n(1)=2n∑r=0n(1)+∑r=0n(1)=(2n)(n+1)+(1)(n+1)=(2n+1)(n+1)\sum_{r=0}^n (2n) + \sum_{r=0}^n (1) = 2n \sum_{r=0}^n (1) + \sum_{r=0}^n (1) = (2n)(n+1) + (1)(n+1) = (2n+1)(n+1)


Of course!
Many thanks :smile:
Reply 7
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Zacken
22
Original post by iMacJack
Of course!
Many thanks :smile:


No worries, good luck for the exam!
Reply 8
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iMacJack
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Original post by Zacken
No worries, good luck for the exam!


Thank you, I'll definitely be needing that luck... haha :smile:

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