Prove by mathematical induction that 11^n-6 is divisible by 5 for all natural numbers n

First I would do the base case (the first value):Test n=1,111-6=5. 5 is divisible by 5 therefore true for n=1.
Now we assume true for n=k,11k-6 is divisible by 5.Next we test n=k+1,11k+1-6We can rearrange this into 1111k-6= 1011k+11k-6We know that for n=k the result is 11k-6 which we assume to be true so that part can be assumed to be true.The first part can be factorised into 5(2*11k) which is divisible by 5. Therefore we have shown that if true for n=k, true for n=k+1 and as we shown true for n=1 it must also be true for all natural numbers. So we have proved this through induction

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Prove by induction that 2^(6n)+3^(2n-2) is divsible by 5. (AS Further pure)


MEI (OCR) M4 June 2006 Q3


A mass m=1kg, initially at rest and with x=10mm, is connected to a damper with stiffness k=24N/mm and damping constant c=0.2Ns/mm. Given that the differential equation of the system is given by d^2x/dt^2+(dx/dt *c/m)+kx/m=0, find the particular solution.


Use De Moivre's Theorem to show that if z = cos(q)+isin(q), then (z^n)+(z^-n) = 2cos(nq) and (z^n)-(z^-n)=2isin(nq).


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