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FP1 OCR June 2014 Q8 Help

I'm doing the OCR June 2014 FP1 paper, but I'm not sure how to go about doing question 8. I know I'm meant to take off the sum of n-1 from 2n from the mark scheme, but I'm not sure why that would work - I know I have to substitute in 2n as n, but I wasn't sure how to get it to start from 1 at the bottom (I don't see how taking away the summation with n-1 as n would work)> Could someone explain how this works?

Question paper:http://www.ocr.org.uk/Images/242444-question-paper-unit-4725-01-further-pure-mathematics-1.pdf

Mark scheme: http://www.ocr.org.uk/Images/235974-mark-scheme-unit-4725-further-pure-mathematics-1-june.pdf
(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by SpaghettiGuy1337
I'm doing the OCR June 2014 FP1 paper, but I'm not sure how to go about doing question 8. I know I'm meant to take off the sum of n-1 from 2n from the mark scheme, but I'm not sure why that would work - I know I have to substitute in 2n as n, but I wasn't sure how to get it to start from 1 at the bottom (I don't see how taking away the summation with n-1 as n would work)Question paper:http://www.ocr.org.uk/Images/242444-question-paper-unit-4725-01-further-pure-mathematics-1.pdfMark scheme: http://www.ocr.org.uk/Images/235974-mark-scheme-unit-4725-further-pure-mathematics-1-june.pdf


Think about the definition of sigma.

For example, let's say I wanted to sum
Unparseable latex formula:

\displaystyle \sum_{r=3}_{6} a_r = a_3 + a_4 + a_5 + a_6

.

We have: ∑r=16ar=a1+a2+a3+a4+a5+a6\displaystyle \sum_{r=1}^{6} a_r = a_1 + a_2 + a_3 + a_4 + a_5 + a_6

And: ∑r=12ar=a1+a2\displaystyle \sum_{r=1}^2 a_r = a_1 + a_2

Subtracting those two:

∑r=16ar−∑r=12ar=(a1+a2+a3+a4+a5+a6)−(a1+a2)=a3+a4+a5+a6=∑r=36ar\displaystyle \sum_{r=1}^{6} a_r - \sum_{r=1}^2 a_r = (a_1 + a_2 + a_3 + a_4 + a_5 + a_6) - (a_1 + a_2) = a_3 + a_4 + a_5 + a_6 = \sum_{r=3}^6 a_r

So, as you can see, if you want to sum between two number ∑r=n2nar\sum_{r=n}^2n a_r, you can instead sum from 1 to 2n and to get a1+⋯+a2na_1 + \cdots + a_{2n} and then subtract away the first 1 to (n-1) terms to get: an+⋯+a2na_n + \cdots + a_{2n} as required.

Hence: ∑r=n2nar=∑r=12nar−∑r=1n−1ar\displaystyle \sum_{r=n}^{2n} a_r = \sum_{r=1}^{2n} a_r - \sum_{r=1}^{n-1} a_r
(edited 7 years ago)
Original post by Zacken
Think about the definition of sigma.

For example, let's say I wanted to sum
Unparseable latex formula:

\displaystyle \sum_{r=3}_{6} a_r = a_3 + a_4 + a_5 + a_6

.

We have: ∑r=16ar=a1+a2+a3+a4+a5+a6\displaystyle \sum_{r=1}^{6} a_r = a_1 + a_2 + a_3 + a_4 + a_5 + a_6

And: ∑r=12ar=a1+a2\displaystyle \sum_{r=1}^2 a_r = a_1 + a_2

Subtracting those two:

∑r=16ar−∑r=12ar=(a1+a2+a3+a4+a5+a6)−(a1+a2)=a3+a4+a5+a6=∑r=36ar\displaystyle \sum_{r=1}^{6} a_r - \sum_{r=1}^2 a_r = (a_1 + a_2 + a_3 + a_4 + a_5 + a_6) - (a_1 + a_2) = a_3 + a_4 + a_5 + a_6 = \sum_{r=3}^6 a_r

So, as you can see, if you want to sum between two number ∑r=n2nar\sum_{r=n}^2n a_r, you can instead sum from 1 to 2n and to get a1+⋯+a2na_1 + \cdots + a_{2n} and then subtract away the first 1 to (n-1) terms to get: an+⋯+a2na_n + \cdots + a_{2n} as required.

Hence: ∑r=n2nar=∑r=12nar−∑r=1n−1ar\displaystyle \sum_{r=n}^{2n} a_r = \sum_{r=1}^{2n} a_r - \sum_{r=1}^{n-1} a_r


That's really well explained. Thank you so much.
Reply 3
Avatar for Zacken
Zacken
22
Original post by SpaghettiGuy1337
That's really well explained. Thank you so much.


Glad it helped, thank you! Appreciated. :smile:

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