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Work Help (Series)

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!
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 (ab)=ab\sum (a-b) = \sum a - \sum b
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
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.
Ahh ok, ill give that a go, I always make that mistake

EDIT - Just thinking, arent they already in brackets which are factorised?
(edited 9 years ago)
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?
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
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?
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
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.
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 :wink: ill just have to re-write it out again
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 :wink: ill just have to re-write it out again


(3n+2)(n1)=3n2n2(3n+2)(n-1)=3n^2-n-2
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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