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Paper 22 math was so hard

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Could you show an example plz? , it would really help (y)
Yeah sure :)

Cubic sequences form: an^3 + bn^2 + cn + d.

We'll take this sequence as an example: 4, 16, 44, 94, 172, 284
Notice that the differences between these terms are: 12, 28, 50, 78, 112. But these differences are not constant. Ergo we will keep going..
Finding the differences once again, we get 16, 22, 28, 34. Yet again they're not constant..
Once more, 6, 6, 6. Now the differences are equal. We stop here. This is the difference at the third level.

Just like how we divide by 2 for the difference in a quadratic sequence, in a cubic we divide by 6. And hence a = (difference at third level / 6) = (6/6) = 1. So now it takes the form:
n^3 + bn^2 + cn + d.

Now is where it gets confusing for most:

We will form 3 equations to obtain the 3 unknowns: 'b', 'c' and 'd'
When n = 1, 1 + b + c + d = 4 [equation A]
n = 2, 8 + 4b + 2c + d = 16 [ B ]
n = 3, 27 + 9b + 3c + d = 44 [C]

From [ B ] - [A] --> 7 + 3b + c = 12 [D]
[C] - [ B ] --> 19 + 5b + c = 28 [E]
[E] - [D] --> 12 + 2b = 16
So b = 2

From [D] 7 + (3*2) + c = 12
So c = - 1

From [A] 1 + 2 - 1 + d = 4
So d = 2

Finally, we can form the rule: (n^3) + (2n^2) - ( n ) +2

If anyone's confused at *any* part, I'll be happy to help =)
 
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Yeah sure :)

Cubic sequences form: an^3 + bn^2 + cn + d.

We'll take this sequence as an example: 4, 16, 44, 94, 172, 284
Notice that the differences between these terms are: 12, 28, 50, 78, 112. But these differences are not constant. Ergo we will keep going..
Finding the differences once again, we get 16, 22, 28, 34. Yet again they're not constant..
Once more, 6, 6, 6. Now the differences are equal. We stop here. This is the difference at the third level.

Just like how we divide by 2 for the difference in a quadratic sequence, in a cubic we divide by 6. And hence a = (difference at third level / 6) = (6/6) = 1. So now it takes the form:
n^3 + bn^2 + cn + d.

Now is where it gets confusing for most:

We will form 3 equations to obtain the 3 unknowns: 'b', 'c' and 'd'
When n = 1, 1 + b + c + d = 4 [equation A]
n = 2, 8 + 4b + 2c + d = 16 [ B ]
n = 3, 27 + 9b + 3c + d = 44 [C]

From [ B ] - [A] --> 7 + 3b + c = 12 [D]
[C] - [ B ] --> 19 + 5b + c = 28 [E]
[E] - [D] --> 12 + 2b = 16
So b = 2

From [D] 7 + (3*2) + c = 12
So c = - 1

From [A] 1 + 2 - 1 + d = 4
So d = 2

Finally, we can form the rule: (n^3) + (2n^2) - ( n ) +2

If anyone's confused at *any* part, I'll be happy to help =)

Thanks alot bro!
 
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I'm getting above 60 in math paper 2 Inshallah
What should I get in paper 42 so that I can guarantee an A*?
 
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My english listening paper sucked!!!! :( I will never get above 30 :( already losing six marks :( I hope the A* range is low this yearrr
 
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But for real i didnt find that Dice question :s
Looking forward to tomorrows Paper 42
Goodluck you guys!!
 
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33
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I'm getting above 60 in math paper 2 Inshallah
What should I get in paper 42 so that I can guarantee an A*?
 
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