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    FP1 induction example

    f(n)=5^{2n+2}-24n-25

    Prove f(n) is divisible by 576 for n=1,2,3,4.....

    f(1)=576

    If (for some k >1) f(k) is divisible by 576 then f(k)=5^{2k+2}-24k-25=576m for some natural number m.

    5^{2k+2}=576m+24k+25

    f(k+1)=5^{2k+4}-24k-49

    f(k+1)=25\times5^{2k+2}-24k-49

    f(k+1)=25(576m+24k+25) -24k-49

    f(k+1)=25\times 576m+25\times 24k+25\times25 -24k-49

    f(k+1)=25\times 576m+24\times 24k+25\times25 -49

    f(k+1)=25\times 576m+576 k+576

    f(k+1)=576 (25m+k+1)

    etc..

    An interesting problem http://www.thestudentroom.co.uk/show....php?t=1293042

    e^{-\frac{1}{2}i\theta}-e^{\frac{1}{2}i\theta}

    =\cos(\frac{1}{2}\theta)-i\sin(\frac{1}{2}\theta)-(\cos(\frac{1}{2}\theta)+i\sin(\  frac{1}{2}\theta))

    =-2i\sin(\frac{1}{2}\theta)
    ____________________
    (1-e^{i\theta})^2=(e^{\frac{1}{2}i\  theta}(e^{-\frac{1}{2}i\theta}-e^{\frac{1}{2}i\theta}))^2=e^{i\  theta}(-2i\sin(\frac{1}{2}\theta))^2=-4e^{i\theta}\sin^2(\frac{1}{2}\t  heta)
    ____________________

    C+iS=1-\binom{2n}{1}(\cos\theta+i\sin\t  heta)+\binom{2n}{2}(\cos2\theta+  i\sin2\theta)-......

    =1-\binom{2n}{1}e^{i\theta}+\binom{  2n}{2}e^{i2\theta}-....=(1-e^{i\theta})^{2n}=(-4e^{i\theta}\sin^2(\frac{1}{2}\t  heta))^n
    =(-4)^n(\cos n\theta+i \sin n\theta)\sin^{2n}(\frac{1}{2}\th  eta)

    C=...
    S=...

    (1-w)^6=(1-e^{i\theta})^6=(-4)^3\cos3\theta\sin^6(\frac{1}{2  }\theta)+i(-4)^3\sin3\theta\sin^6(\frac{1}{2  }\theta)

    If w is a cube root of 1 then theta=0, 2pi/3, 4pi/3

    (1-w)^6 = -27, 0, 27
    _______________________________

    Step 1 2010 Q13

    P(first text in second hour) = P(0 in first hour)*P(>0 in second hour)

    p=e^{-\lambda}(1-e^{-\lambda})

    gives the required equation pe^{2\lambda}-e^{\lambda}+1=0

    For the two positive values of lambda.. one is easy, for the other show that e^{\lambda}=\frac{1}{2p}(1-\sqrt{1-4p})&gt;1 for 0<p<1/4

    Starting with 1-4p+4p^2&gt;1-4p leads to \frac{1-\sqrt{1-4p}}{2p}&gt;1 so e^{\lambda}&gt;1 and so \lambda&gt;0

    Mildred's lamdas satisfy pe^{2\lambda}-e^{\lambda}+1=0

    so using product of roots e^{\lambda_1+\lambda_2}=\frac{1}  {p}

    \lambda_1+\lambda_2=-\ln p

    The combined rate for the two phones is \lambda_1+\lambda_2 subbing this in for lambda in the given equation results in p(1-p) as the required probability.
 
 
 
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Updated: July 11, 2009

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