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FP1: SIGMA summations: Simplifying common factors

Question 4)a).

Question Paper:
http://filestore.aqa.org.uk/subjects/AQA-MFP1-QP-JAN12.PDF

Mark Scheme:
http://filestore.aqa.org.uk/subjects/AQA-MFP1_W-MS-JAN12.PDF

I have to self-teach myself this topic.

I don't understand how you take out common factors and combine them. Can someone teach me how you do these kinds of questions assuming that I only have enough knowledge to do C1 and C2?

I substituted the formula and get:
4(0.25)n^2(n+1)^2 - 3(1/6)n(n+1)(2n+1)

I simplify it to
n^2(n+1)^2 - (1/2)n(n+1)(2n+1)

Now what do I do? I have no idea.
In the mark scheme it says "take out common factors n and (n + 1)" but how? I don't understand what that means.
Reply 1
Factor out the common factor from the two terms in the difference.

n(n+1)[n(n+1)]n(n+1)[12(2n+1)]n(n+1)[n(n+1)] - n(n+1)[\frac{1}{2}(2n+1)]

Do you see the common factor now?
(edited 10 years ago)
Reply 2
Yup, the common factor is n(n+1).

So, how is the simplification process done?

And how do you write text like that?

I have self-taught many chapters of FP1 but I just don't understand this bit.
Reply 3
Original post by Konnichiwa
Question 4)a).

Question Paper:
http://filestore.aqa.org.uk/subjects/AQA-MFP1-QP-JAN12.PDF

Mark Scheme:
http://filestore.aqa.org.uk/subjects/AQA-MFP1_W-MS-JAN12.PDF

I have to self-teach myself this topic.

I don't understand how you take out common factors and combine them. Can someone teach me how you do these kinds of questions assuming that I only have enough knowledge to do C1 and C2?

I substituted the formula and get:
4(0.25)n^2(n+1)^2 - 3(1/6)n(n+1)(2n+1)

I simplify it to
n^2(n+1)^2 - (1/2)n(n+1)(2n+1)

Now what do I do? I have no idea.
In the mark scheme it says "take out common factors n and (n + 1)" but how? I don't understand what that means.


What do you mean when you say you do not understand

Could you factorise a2babca^2b - abc for example?
(edited 10 years ago)
Reply 4
Original post by TenOfThem
What do you mean when you say you do not understand

Could you factorise a2babca^2b - abc for example?

a and b are a factor, I think I understand now, that "a^2b - abc" has now made me figure it out what it meant! Thanks so much.
ab(a - c)
Reply 5
Original post by Konnichiwa
Yup, the common factor is n(n+1).

So, how is the simplification process done?

And how do you write text like that?

I have self-taught many chapters of FP1 but I just don't understand this bit.


Just use [ latex] tags, or look at the Latex sticky in this board.

Anyways, it's not exactly "simplification", because it can be argued either way which form is simpler.

The thing is that they ask for the answer in the form:

kn(n+1)(2n21)kn(n+1)(2n^2-1)

Which means that you want to turn it into the product of 4 factors, n(n+1) being one of them. So it's not exactly simplification, but guided factorization.
(edited 10 years ago)
Reply 6
Original post by Konnichiwa
a and b are a factor, I think I understand now, that "a^2b - abc" has now made me figure it out what it meant! Thanks so much.
ab(a - c)


:smile:

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