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    (Original post by Nikhilm)
    For proof by induction, what does all the notation (e.g. nEZ+) actually mean?
    It means that n is a member of the set of positive real numbers {1,2,3,...}
    edit: integers not just reals lmao
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    (Original post by Nikhilm)
    For proof by induction, what does all the notation (e.g. nEZ+) actually mean?
    n is a positive integer
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    Anyone know where I have gone wrong?
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    (Original post by wr123)
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    Anyone know where I have gone wrong?
    You have not gone wrong however there is another step you need to do. Think about seperating the terms so that you have 4f(k) + ....
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    (Original post by wr123)
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    Anyone know where I have gone wrong?
    You haven't.

    You have:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 3(5^k) + 5^k + 2(11^k) + 18(11^k) \\&= 3(5^k + 6(11^k)) + f(k)\end{align*}

    Which is all clearly divisible by 3.

    Or alternatively:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 4(5^k) + 8(11^k) + 12(11^k) \\ & = 4(5^k + 2(11^k)) + 3(4(11^{k})) \\ & = 4f(k) + 3 \cdot 4\cdot 11^k   \end{align*}
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    (Original post by Zacken)
    You haven't.

    You have:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 3(5^k) + 5^k + 2(11^k) + 18(11^k) \\&= 3(5^k + 6(11^k)) + f(k)\end{align*}

    Which is all clearly divisible by 3.

    Or alternatively:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 4(5^k) + 8(11^k) + 12(11^k) \\ & = 4(5^k + 2(11^k)) + 3(4(11^{k})) \\ & = 4f(k) + 3 \cdot 4\cdot 11^k   \end{align*}
    (Original post by alfmeister)
    You have not gone wrong however there is another step you need to do. Think about seperating the terms so that you have 4f(k) + ....
    (Original post by Zacken)
    You haven't.

    You have:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 3(5^k) + 5^k + 2(11^k) + 18(11^k) \\&= 3(5^k + 6(11^k)) + f(k)\end{align*}

    Which is all clearly divisible by 3.

    Or alternatively:

    \displaystyle 

\begin{align*}4(5^k) + 20(11^k) &= 4(5^k) + 8(11^k) + 12(11^k) \\ & = 4(5^k + 2(11^k)) + 3(4(11^{k})) \\ & = 4f(k) + 3 \cdot 4\cdot 11^k   \end{align*}
    Thank you both! I just couldn't see it!
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    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    For this paper- Q 5(b)- Why is it Z=YX^(-1) and NOT Z=X^(-1)Y??

    Thanks!!

    Edit: This is my working btw:
    ZX=Y
    ZX^(-1)X=X^(-1)Y
    Z=X^(-1)Y
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    Hi everyone,

    Here are the first set of solutions to the practice paper of FP1 questions found on page 24 of this forum thread.
    Attached Files
  1. File Type: docx FP1 Doc 2.docx (11.4 KB, 62 views)
  2. File Type: docx FP1 Doc 2.docx (706.2 KB, 56 views)
  3. File Type: docx FP1 Doc 2.docx (678.4 KB, 56 views)
  4. File Type: docx FP1 Doc 2.docx (657.2 KB, 45 views)
  5. File Type: docx FP1 Doc 2.docx (630.1 KB, 54 views)
  6. File Type: docx FP1 Doc 2.docx (612.2 KB, 55 views)
  7. File Type: docx FP1 Doc 2.docx (666.9 KB, 42 views)
  8. File Type: docx FP1 Doc 2.docx (624.4 KB, 40 views)
  9. File Type: docx FP1 Doc 2.docx (615.5 KB, 46 views)
  10. File Type: docx FP1 Doc 2.docx (641.4 KB, 48 views)
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    (Original post by thesmallman)
    https://7cba9babeb0db0ff9468853e0b2d...%20Edexcel.pdf

    For this paper- Q 5(b)- Why is it Z=YX^(-1) and NOT Z=X^(-1)Y??

    Thanks!!

    Edit: This is my working btw:
    ZX=Y
    ZX^(-1)X=X^(-1)Y
    Z=X^(-1)Y
    I think its because you're post multiplying by X^-1 to get rid of it on the LHS you do the same on the RHS.

    If it were XZ=Y then you'd (pre) multiply by X^-1 So X^-1Z=X^-1Y
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    Here are the second set of solutions.
    Attached Files
  11. File Type: docx FP1 Doc 2.docx (578.5 KB, 45 views)
  12. File Type: docx FP1 Doc 2.docx (641.9 KB, 40 views)
  13. File Type: docx FP1 Doc 2.docx (635.3 KB, 37 views)
  14. File Type: docx FP1 Doc 2.docx (666.1 KB, 43 views)
  15. File Type: docx FP1 Doc 2.docx (647.6 KB, 38 views)
  16. File Type: docx FP1 Doc 2.docx (645.5 KB, 42 views)
  17. File Type: docx FP1 Doc 2.docx (659.7 KB, 35 views)
  18. File Type: docx FP1 Doc 2.docx (667.8 KB, 41 views)
  19. File Type: docx FP1 Doc 2.docx (614.9 KB, 43 views)
  20. File Type: docx FP1 Doc 2.docx (687.8 KB, 42 views)
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    Posted from TSR Mobile
    There you go, sorry if my wording is not the best but I tried to explain. Name:  1463583970576.jpg
Views: 84
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    Finally here is the third set of solutions.
    Attached Files
  21. File Type: docx FP1 Doc 2.docx (656.3 KB, 48 views)
  22. File Type: docx FP1 Doc 2.docx (650.6 KB, 43 views)
  23. File Type: docx FP1 Doc 2.docx (677.2 KB, 36 views)
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    (Original post by Starlight7738)
    Finally here is the third set of solutions.
    Cheers-am doing your paper atm
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    (Original post by SB0073)
    I think its because you're post multiplying by X^-1 to get rid of it on the LHS you do the same on the RHS.

    If it were XZ=Y then you'd (pre) multiply by X^-1 So X^-1Z=X^-1Y
    hmmm I think I get where youre coming from but I'll try to use some examples to confirm this, cheers
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    (Original post by SB0073)
    I think its because you're post multiplying by X^-1 to get rid of it on the LHS you do the same on the RHS.

    If it were XZ=Y then you'd (pre) multiply by X^-1 So X^-1Z=X^-1Y
    yep- can confirm that you're correct. Wow glad I found out about this be4 the exam aha
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    For the summation stuff (not mathematical induction), they always say, 'show that ... for all positive integers of n'. Can it ever be 'for all n' and if so how would you go about solving that?
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    (Original post by Nikhilm)
    For the summation stuff (not mathematical induction), they always say, 'show that ... for all positive integers of n'. Can it ever be 'for all n' and if so how would you go about solving that?
    No.
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    How is 4b 8? in jan 2016 ial
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    (Original post by LLk1)
    How is 4b 8? in jan 2016 ial
    lol talk about pulling numbers out of your arse, they got 8 from absolutey nowhere! - legit stuck on that qn for a good 5 mins and eventually gave up. 1 mark so meh
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    can someone check this method for me

    jan 16 ial

    qn 9 a - i managed to factor out a 21 using the method where you equate the divisor to f(k) ie 21m = f(k)

    i got 21 ( 21m - 4.4^k)
    ie 21 ( f(k) - 4.4^k )
    is that alright to leave it like that - MS had it differently ...
 
 
 
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