Work Help (Series) Watch

PanthersWill
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#1
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#1
Been doing some work and my brain has gone to mush on 1 question, I seem to be around half way through proving it, but not sure where to go or if i have gone wrong, the question is below, is anyone is able to show me how they would do it (maybe by writing it out then taking a picture) or typing it which ever is easiest. Then i would be very grateful thanks!
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Mr M
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#2
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(Original post by PanthersWill)
Been doing some work and my brain has gone to mush on 1 question, I seem to be around half way through proving it, but not sure where to go or if i have gone wrong, the question is below, is anyone is able to show me how they would do it (maybe by writing it out then taking a picture) or typing it which ever is easiest. Then i would be very grateful thanks!
We are not going to do your work for you.

Expand the bracket.

Recall that \sum (a-b) = \sum a - \sum b
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PanthersWill
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#3
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#3
I already have done that, like i said, im halfway through but cant seem to get further, I think i could somehow get to the answer, but it means expanding all brackets and getting a cubic equation but then it means having to solve that as well, which doesnt seem the right way to go about it
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Mr M
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#4
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(Original post by PanthersWill)
I already have done that, like i said, im halfway through but cant seem to get further, I think i could somehow get to the answer, but it means expanding all brackets and getting a cubic equation but then it means having to solve that as well, which doesnt seem the right way to go about it
You shouldn't expand. FACTORISE.
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PanthersWill
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#5
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Ahh ok, ill give that a go, I always make that mistake

EDIT - Just thinking, arent they already in brackets which are factorised?
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Mr M
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(Original post by PanthersWill)
Ahh ok, ill give that a go, I always make that mistake

EDIT - Just thinking, arent they already in brackets which are factorised?
So which are the common factors?
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PanthersWill
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#7
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#7
Im just going round in circles, getting answers close but not correct and cant spot where i should do something different :/ Am i right saying n+1 will be common factor?
What i am doing is taking 1/12 n outside the equation so it leaves 3n(n+1)(n+1)-2(n+1)(2n+1). then i take n+1 out as well? so 1/12 n(n+1)[3n(n+1)-2(2n+1)] then this is where i get stuck and end up getting a wrong answer, have i done anything wrong already which will lead to the wrong answer or am i on right path (if so, how would i tackle the next step) as clearly (n+1) isnt in the answer
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Mr M
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#8
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(Original post by PanthersWill)
Im just going round in circles, getting answers close but not correct and cant spot where i should do something different :/ Am i right saying n+1 will be common factor?
What i am doing is taking 1/12 n outside the equation so it leaves 3n(n+1)(n+1)-2(n+1)(2n+1). then i take n+1 out as well? so 1/12 n(n+1)[3n(n+1)-2(2n+1)] then this is where i get stuck and end up getting a wrong answer, have i done anything wrong already which will lead to the wrong answer or am i on right path (if so, how would i tackle the next step) as clearly (n+1) isnt in the answer
That's correct. Why do you think (n+1) can't be in the final answer?
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PanthersWill
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#9
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#9
Well i have to show that it gets 1/12n(n^2-1)(3n+2). I do believe it could be (n+1) and (n-1) to get the n^2-1. But im not sure how i get there? do i then expand then 3n(n+1)-2(2n+1)? to get 3n^2-n-2? or not
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Mr M
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#10
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(Original post by PanthersWill)
Well i have to show that it gets 1/12n(n^2-1)(3n+2). I do believe it could be (n+1) and (n-1) to get the n^2-1. But im not sure how i get there? do i then expand then 3n(n+1)-2(2n+1)? to get 3n^2-n-2? or not
I don't really like the answer you think you are trying to achieve as it isn't the simplest factorised form. But yes all of that is right.
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PanthersWill
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#11
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I know what you mean, but thats what the question asks haha i must have gone wrong in final working as i ended with 3n-2 and n+1, when the signs are wrong. I dont know what i did but i can see that the signs should be other way round otherwise when expanded it doesnt work ill just have to re-write it out again
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Mr M
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#12
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#12
(Original post by PanthersWill)
I know what you mean, but thats what the question asks haha i must have gone wrong in final working as i ended with 3n-2 and n+1, when the signs are wrong. I dont know what i did but i can see that the signs should be other way round otherwise when expanded it doesnt work ill just have to re-write it out again
(3n+2)(n-1)=3n^2-n-2
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physicsmaths
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#13
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#13
the way i used to do these was by factoring the outside thing, in this case 1/12 slmething then simplify the inside. worked every time.


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