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FP1 help! Series

Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

Thank you!
Reply 1
Original post by iMacJack
Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

Thank you!


You'll need to use the result that you proved in the first part, so it'd be helpful if you could show us that.
Original post by iMacJack
Hi guys, I am just doing a past paper for FP1 and have stumbled across this sort of question a few times and I am unable to answer it! Could you please give me a little bit of assistance as to what to do please?

Thank you!


I got the answer to be 2213/29, could you check if this is right then I can explain my working if you would like :smile:
Reply 3
Original post by Dapperblook22
I got the answer to be 2213/29, could you check if this is right then I can explain my working if you would like :smile:


It's wrong.
Reply 4
Oh, I remember this question... it was in my FP1 exam. :tongue:
Original post by Zacken
It's wrong.


Oops, that's why I asked if it was right :tongue:. I would assume an arithmetic error, as I split the problem up to 8 times the sum of r^3 from 5 to 8, minus 3 times the sum of r^2 from 5 to 8 plus K times the sum of r from 5 to 8
Reply 6
This is pretty easy, just use the standard series formulae, and basic algebra, to find k
Reply 7
Original post by Dapperblook22
Oops, that's why I asked if it was right :tongue:. I would assume an arithmetic error, as I split the problem up to 8 times the sum of r^3 from 5 to 8, minus 3 times the sum of r^2 from 5 to 8 plus K times the sum of r from 5 to 8


The sums are to 10 not 8. The correct answer is

Spoiler



The way you did it is good, but the first part of this question makes this part a trivial 2 liner or such instead of having to manually re-derive everything again.
Reply 8
Hey guys - sorry!
Here's the full question
https://gyazo.com/8024cf8825c10420ba3c3a62b256dbee

Appreciate the help you guys are going to give :smile:
Original post by Zacken
The sums are to 10 not 8. The correct answer is

Spoiler


The way you did it is good, but the first part of this question makes this part a trivial 2 liner or such instead of having to manually re-derive everything again.


Ah, I might need glasses, honestly thought it was an 8 :tongue:. At least the method was correct so OP may follow that if they want.
Reply 10
Original post by iMacJack
Hey guys - sorry!
Here's the full question
https://gyazo.com/8024cf8825c10420ba3c3a62b256dbee

Appreciate the help you guys are going to give :smile:


So right away you can split your sum as r=510(8r33r)+r=510kr2\displaystyle \sum_{r=5}^{10} (8r^3 - 3r) + \sum_{r=5}^{10} kr^2

You should be able to say that, right away:

r=510(8r33r)=r=110(8r33r)r=14(8r33r)\displaystyle \sum_{r=5}^{10} (8r^3 - 3r) = \sum_{r=1}^{10} (8r^3 - 3r) - \sum_{r=1}^{4} (8r^3 - 3r)

which lets you compute that by plugging in the relevant values of nn in the formula derived in the first part.

Similarly: kr=510kr2=k(r=110r2r=14r2)k\sum_{r=5}^{10} kr^2 = k\left(\sum_{r=1}^{10} r^2 - \sum_{r=1}^{4} r^2 \right) where standard formula booklet plugging in evalutes the bracket.

This gets you 355k+some number=that number I don’t remember355k + \text{some number} = \text{that number I don't remember} and you solve for kk.
Reply 11
Original post by Zacken
So right away you can split your sum as r=510(8r33r)+r=510kr2\displaystyle \sum_{r=5}^{10} (8r^3 - 3r) + \sum_{r=5}^{10} kr^2

You should be able to say that, right away:

r=510(8r33r)=r=110(8r33r)r=14(8r33r)\displaystyle \sum_{r=5}^{10} (8r^3 - 3r) = \sum_{r=1}^{10} (8r^3 - 3r) - \sum_{r=1}^{4} (8r^3 - 3r)

which lets you compute that by plugging in the relevant values of nn in the formula derived in the first part.

Similarly: kr=510kr2=k(r=110r2r=14r2)k\sum_{r=5}^{10} kr^2 = k\left(\sum_{r=1}^{10} r^2 - \sum_{r=1}^{4} r^2 \right) where standard formula booklet plugging in evalutes the bracket.

This gets you 355k+some number=that number I don’t remember355k + \text{some number} = \text{that number I don't remember} and you solve for kk.

Thank you mate. I will give this a go now and let you know how it goes!
Thanks :smile: <3
Reply 12
Original post by iMacJack
Thank you mate. I will give this a go now and let you know how it goes!
Thanks :smile: <3


No problem, this method of splitting sums from one number to another by writing them as sums from 1 to a number - sum from 1 to another number is in pretty much every FP1 paper ever.
Original post by iMacJack
Thank you mate. I will give this a go now and let you know how it goes!
Thanks :smile: <3


If you need the working, please find it attached. Zacken explained it pretty clearly so you may not even need this :tongue:

Also sorry if it is a bit choppy at places.

IMG_1359[1].jpg
Reply 14
I got K = -497/355 which = -7/5??
Reply 15
Original post by Zacken
No problem, this method of splitting sums from one number to another by writing them as sums from 1 to a number - sum from 1 to another number is in pretty much every FP1 paper ever.


yeah I can do the normal ones but when there's a constant involved it seems to throw me off a bit! thanks for the help!
Reply 16
Original post by iMacJack
yeah I can do the normal ones but when there's a constant involved it seems to throw me off a bit! thanks for the help!


Constants are constants, pull them out and do your thang with the series.
Reply 17
Original post by Dapperblook22
If you need the working, please find it attached. Zacken explained it pretty clearly so you may not even need this :tongue:

Also sorry if it is a bit choppy at places.

IMG_1359[1].jpg


Yeah thank you, appreciated :smile:

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