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    What I did was -


    U(n) = 7(7^2n) + 125(5^n)

    U(n + 1) = 343(7^2n) + 625(5^n)

    U(n + 1) - 6U(n) = 308(7^2n)

    So as 308 is divisible by 44, U(n) is divisible by 44.

    This question was in Further Maths (9231) winter 2010 paper 1.

    In the mark scheme of this paper, there is a different way of solving this which I was not quite able to understand.

    Here is the mark scheme -



    Is my solution correct? and can someone explain me the solution in the mark scheme!
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    (Original post by Rishabh95)


    What I did was -


    U(n) = 7(7^2n) + 125(5^n)

    U(n + 1) = 343(7^2n) + 625(5^n)

    U(n + 1) - 6U(n) = 308(7^2n)

    So as 308 is divisible by 44, U(n) is divisible by 44.

    This question was in Further Maths (9231) winter 2010 paper 1.

    In the mark scheme of this paper, there is a different way of solving this which I was not quite able to understand.

    Here is the mark scheme -



    Is my solution correct? and can someone explain me the solution in the mark scheme!
    Your solution is also correct (apart from the fact that you wrote U(n + 1) - 6U(n), rather than U(n + 1) - 5U(n), but I presume that's a typo).

    I would double check how your exam board wants you to lay out the answer to these induction questions as I can't really see the usual structure in your solution. You may have typed it out in such a way to save space, I'm not too sure.

    In the first solution they give, they assume that U(k) is divisible by 44 and then proceed to take the inductive step by considering U(k+1). In terms of manipulation, what they did was:
    U(k+1) = 7^{2(k+1)+1} + 5^{(k+1)+3}
    =7^{(2k+1)+2} + 5^{(k+3)+1}
    =7^2 \times 7^{2k+1} + 5 \times 5^{k+3}
    =49 \times 7^{2k+1} + (49 \times 5^{k+3} - 44\times 5^{k+3})
    =49 \times (7^{2k+1} + 5^{k+3}) - 44\times 5^{k+3}
    =49U(k) - 44\times 5^{k+3}

    Since U(k) is divisible by 44 (as in our assumption) and obviously 44 is, both terms are divisible by 44 and thus is the whole expression for U(k+1). Hence it follows that U(n) is divisible by 44.

    They use a very similar argument for the alternative solution, except they subtract U(k) from the start and then manipulate.
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    (Original post by Farhan.Hanif93)
    Your solution is also correct (apart from the fact that you wrote U(n + 1) - 6U(n), rather than U(n + 1) - 5U(n), but I presume that's a typo).

    I would double check how your exam board wants you to lay out the answer to these induction questions as I can't really see the usual structure in your solution. You may have typed it out in such a way to save space, I'm not too sure.

    In the first solution they give, they assume that U(k) is divisible by 44 and then proceed to take the inductive step by considering U(k+1). In terms of manipulation, what they did was:
    U(k+1) = 7^{2(k+1)+1} + 5^{(k+1)+3}
    =7^{(2k+1)+2} + 5^{(k+3)+1}
    =7^2 \times 7^{2k+1} + 5 \times 5^{k+3}
    =49 \times 7^{2k+1} + (49 \times 5^{k+3} - 44\times 5^{k+3})
    =49 \times (7^{2k+1} + 5^{k+3}) - 44\times 5^{k+3}
    =49U(k) - 44\times 5^{k+3}

    Since U(k) is divisible by 44 (as in our assumption) and obviously 44 is, both terms are divisible by 44 and thus is the whole expression for U(k+1). Hence it follows that U(n) is divisible by 44.

    They use a very similar argument for the alternative solution, except they subtract U(k) from the start and then manipulate.
    Yeah, it was a typo! and thanks for the help!
 
 
 
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