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    Let  u_n = \frac{6(n-1)}{(n-2)(n-3)}.

    How do you tell that  u_{n+1} < u_n ?

    Thanks.
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    What's u_{n+1} ?

    Then show that u_{n+1}-u_{n} < 0
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    (Original post by notnek)
    What's u_{n+1} ?

    Then show that u_{n+1}-u_{n} < 0
     u_{n+1} - u_n = \frac{6n}{(n-1)(n-2)} - \frac{6(n-1)}{(n-2)(n-3)}
    =  \frac{6n(n-3)-6(n-1)^{2}}{(n-1)(n-2)(n-3)}
    =  \frac{-6(n+1)}{(n-1)(n-2)(n-3)}

    How do you show this is less than 0?

    By the way, I think of other method, in which I attempt to show that, if  u_{n+1} < u_n , then  \frac{u_{n+1}}{u_n} < 1 . But I found it difficult to prove as well.
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    just a quick question.. which AS/A2 Level Module is this in?
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    Well for n > 3 the top is < 0, and the bottom is > 0.

    Since u_n isn't defined for n=2,3 and u_n+1 isn't defined for n=1, I think you can reasonably assume n > 3 here.
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    (Original post by mfc20)
    just a quick question.. which AS/A2 Level Module is this in?
    This is stated at the solution of an AEA question, in which it states that  k_{n+1} &lt; k_n without proof.

    (Original post by DFranklin)
    Well for n > 3 the top is < 0, and the bottom is > 0.

    Since u_n isn't defined for n=2,3 and u_n+1 isn't defined for n=1, I think you can reasonably assume n > 3 here.
    I'm really confused at those things, but can n be negative integers?

    Thanks.
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    It can, if you defined the sequence for negative indices.
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    what you need to note is that as you get large n the denominator will increase faster than the numerator, so since n_x is going to be a certain number y the n_x+1 will be even smaller than n_x since it is a decreasing function.

    don't you have to show that it gives integer solutions for n=something anyways?
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    (Original post by Civ-217)
    I'm really confused at those things, but can n be negative integers?
    A sequence is a function which has positive integers as its "input" (called the domain in C2), so n cannot take negative integer values. In other words, the function, denoted by u_n here is  f(n) = u_n =  \frac{6(n-1)}{(n-2)(n-3)}, and the input/domain is all the positive integers which make sense. Because the values n =2 and n=3 give division by zero, which we cannot have, we have to conclude that the sequence u_n is not defined for those two values of n.
 
 
 

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