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# MAT Prep Thread - 2nd November 2016

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1. (Original post by KloppOClock)
Hope you're okay, still plenty of time to get prepared for it.
Thanks for the concern, I'm doing alright just a bit drowsy from the anaesthetic but will be fine. didn't feel good for the three weeks prior to today. That is true ,Ill just have to get my head down over the next few days and try forget about the pain.
2. What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

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3. For 2011 Q5 ii can someone tell me if n^2-3n+4 would be an acceptable answeer because its not the answer they have given but I provided working and it seems to work ?
4. (Original post by theaverage)
What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

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The recurrence is a lot easier to understand in terms of m_n = l_n + 2.

Have you looked at the solution?
5. (Original post by Mystery.)
For 2011 Q5 ii can someone tell me if n^2-3n+4 would be an acceptable answeer because its not the answer they have given but I provided working and it seems to work ?
Do you mean 2011 Q5 (ii)? That is not asking for a specific answer. Part (iii) is, but the given answer of 2^(n-1) is very different from yours.
6. (Original post by theaverage)
What did everyone do for part iii question 5 2012?

I don't see how they used the hint?

Posted from TSR Mobile
I didn't get the same answer as them but I got a different answer that works, I don't get how they used the hint either.
7. (Original post by RichE)
Do you mean 2011 Q5 (ii)? That is not asking for a specific answer. Part (iii) is, but the given answer of 2^(n-1) is very different from yours.
Yeah iii I meant. I know its different but my reasoning is that there is only 1 way to get to the first shaded and last shaded square. so that's 2 then for each of the left over squares there are n-1 ways to get to it, therefore (n-1)(n-2) + 2 ?
8. (Original post by Mystery.)
Yeah iii I meant. I know its different but my reasoning is that there is only 1 way to get to the first shaded and last shaded square. so that's 2 then for each of the left over squares there are n-1 ways to get to it, therefore (n-1)(n-2) + 2 ?
Yours can't be the right answer as it's a different function.

If you wanted to argue it your way you would be summing binomial coefficients

1 + (n-1)C1 + (n-2)C2 + ... + (n-1)C(n-1) + 1

which would again give their answer of 2^(n-1).
9. (Original post by RichE)
Yours can't be the right answer as it's a different function.

If you wanted to argue it your way you would be summing binomial coefficients

1 + (n-1)C1 + (n-2)C2 + ... + (n-1)C(n-1) + 1

which would again give their answer of 2^(n-1).
I know its not the right answer obviously but it works. I am not saying its the same
10. (Original post by Mystery.)
I know its not the right answer obviously but it works. I am not saying its the same
In what sense does it work? Your reasoning is wrong (see my previous post) and your formula gives the wrong answers.
11. (Original post by RichE)
In what sense does it work? Your reasoning is wrong (see my previous post) and your formula gives the wrong answers.
well when I tried it with different n x n grids it gave the same answer as their function.
12. (Original post by Mystery.)
well when I tried it with different n x n grids it gave the same answer as their function.

(n-1)(n-2) + 2

2^(n-1).

The second function grows exponentially and the first is a polynomial, so there's no way they could be the same - for large n the second function would be colossally bigger.

But you only have to try n = 5 to get different answers from the two formulae - 14 from the first and 16 from the second. (Try n=100 on your calculator to see how different the formulae really are.)
13. Hi
In MAT 2007, Q2, part (ii) where we find an expression in terms of k....

How would you come up with (4k - 1) / 3 as the last part of that expression? Is that just pattern spotting or is there a method to develop a relationship between k and the final term of each expression in the sequence?
Hi
In MAT 2007, Q2, part (ii) where we find an expression in terms of k....

How would you come up with (4k - 1) / 3 as the last part of that expression? Is that just pattern spotting or is there a method to develop a relationship between k and the final term of each expression in the sequence?
It's a geometric sum.
15. Can anyone explain Part V, question 5 of the 2007 paper?
Thanks
16. (Original post by RichE)
It's a geometric sum.
Ah yes, can't believe I didn't spot that...
But how do we take into account the "+1" for each term? As in for the geometric series... r is 4 but don't we add 1 to each term to get the next term also?
Ah yes, can't believe I didn't spot that...
But how do we take into account the "+1" for each term? As in for the geometric series... r is 4 but don't we add 1 to each term to get the next term also?
I'm not really sure what you're asking. The (4^k-1)/3 expression appears as it's the sum of the geometric series given.

That multiplying by 4 and adding 1 each time leads to a geometric sum at all is just a matter of spotting the pattern

1
4+1
4^2 + 4 + 1
4^3 + 4^2 + 4 + 1
18. What are people generally getting for the 2009 and 2011 papers?
19. What are the best resources for improving at MAT papers, I feel like Im not really understanding how to approach most of these questions
20. Do you guys normally do all the questions or do like 2/3 from Q2-3 and try to secure all the marks?

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