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Fp1 series quick question

Screenshot_20160516-200342.jpg

I understand how to expand r squared and 2r.

But I'm stuck on 2n + 1.

In the mark scheme they have done this: 2n(n+1) and (n+1) so it looks like 2n and 1 have been multiplied by (n+1)...could someone explain the understanding behind this please?
Reply 1
Avatar for Zacken
Zacken
22
Original post by Ravster
In the mark scheme they have done this: 2n(n+1) and (n+1) so it looks like 2n and 1 have been multiplied by (n+1)...could someone explain the understanding behind this please?


This is the 6th time I'm answering this question on TSR, seems like a good question! :lol:

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

1. ∑r=0n(2n+1)=2(n)+1⏟0th   term+∑r=1n(2n+1)=(2n+1)+(2n+1)n=(2n+1)(n+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. ∑r=0nf(r)=f(0)+f(1)+f(2)+⋯+f(n)⏟n terms⏞(n+1) terms\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 ∑r=0n(2n+1)=(2n+1)∑r=0n1\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

Then you have f(r)=1f(r) = 1 being added together n+1n+1 times to get a total of (2n+1)(1+1+⋯+1)=(2n+1)(n+1)(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1).
What paper year is this? I've never seen anything like this before, looks hard!
Reply 3
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Zacken
22
Original post by 16characterlimit
What paper year is this? I've never seen anything like this before, looks hard!


It's not any different to a normal series question.
Original post by 16characterlimit
What paper year is this? I've never seen anything like this before, looks hard!


June 2013 (R) and you're actually quite right. This was the only question of its kind in all the past papers, in the sense that it started from r=0, not hard, just different to the normal questions
Original post by Zacken
It's not any different to a normal series question.


But you are summing r normally, this one you have to sum n.

edit: NEVER MIND, I haven't done FP1 for a while.

edit2: no I am confused now, the standard sums only apply for the variable below the sum, r in this case. Help?
(edited 7 years ago)
Reply 6
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Zacken
22
Original post by 16characterlimit
But you are summing r normally, this one you have to sum n.

edit: NEVER MIND, I haven't done FP1 for a while.

edit2: no I am confused now, the standard sums only apply for the variable below the sum, r in this case. Help?


Look at my post, nn is just some random constant.

So ∑r=1n(2n+1)=(2n+1)∑r=1n(1)\sum_{r=1}^n (2n+1) = (2n+1)\sum_{r=1}^n (1) and you can't really get a simpler sum than just summing 1's up...
Original post by Zacken
Look at my post, nn is just some random constant.

So ∑r=1n(2n+1)=(2n+1)∑r=1n(1)\sum_{r=1}^n (2n+1) = (2n+1)\sum_{r=1}^n (1) and you can't really get a simpler sum than just summing 1's up...


Oh right, thanks crises averted.

Isn't a bit strange using the same letter for the constant as in the sum, like saying differentiate x + x, where the second x is a constant? Or am I missing something important.
Reply 8
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Zacken
22
Original post by 16characterlimit
Oh right, thanks crises averted.

Isn't a bit strange using the same letter for the constant as in the sum, like saying differentiate x + x, where the second x is a constant? Or am I missing something important.


No, it's not. It's like you saying, "heeey! You're not allowed to do ∑r=15(5)\sum_{r=1}^5 (5) because the number at the top and in the sum are the same wtf?!??!?!?!?!/!?!?". nn is a fixed unchanging constant. rr is the only thing that's changing.
Original post by Zacken
No, it's not. It's like you saying, "heeey! You're not allowed to do ∑r=15(5)\sum_{r=1}^5 (5) because the number at the top and in the sum are the same wtf?!??!?!?!?!/!?!?". nn is a fixed unchanging constant. rr is the only thing that's changing.


thx +rep
Reply 10
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Zacken
22
Original post by 16characterlimit
thx +rep


No prob. :smile:
What exam board and year paper is this question from?
Reply 12
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Zacken
22
Original post by Bobunitrd
What exam board and year paper is this question from?


Edexcel June 2013 (R), as mentioned above in the thread.
Original post by Zacken
Edexcel June 2013 (R), as mentioned above in the thread.


Oh sorry didn't see that, thanks anyway
Reply 14
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Zacken
22
Original post by Bobunitrd
Oh sorry didn't see that, thanks anyway


No problem.

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