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    1). Prove by induction, for all positive integers n, that:

    (1 x 5) + (2 x 6) + (3 x 7) + ..... + n(n + 4) = 1/6 n(n+1)(2n+13)


    2). A sequence U1, U2, U3.....Un is defined by Un+1 = (3Un - 1)/4

    The first term (U1) = 2

    Prove by induction that Un = 4 [(3/4)^n] - 1


    Help?!?
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    what have you done so far?
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    for question 2,

    ive proved it for n = 1,

    and now I have to prove it for n + 1

    so ive just written it out:

    4 [(3/4)^n+1] - 1

    But what do I do now?
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    anyone?
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    (Original post by DeanK2)
    what have you done so far?
    any help ?
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    You need to be careful in setting out proofs by induction. So it's true for n = 1, and hence for some n = k, U_k = 4 \left( \dfrac{3}{4} \right)^k - 1.

    So U_{k+1} = \dfrac{3U_k - 1}{4}, and substitute 4 \left( \dfrac{3}{4} \right)^k - 1 in place of U_k.
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    which one do you need help with?
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    4 [(3/4)^n+1] - 1 = Un+1


    so now separte Un+1 into Un + (n+1) ie the nth term + the n+first term

    so
    Un+1 = Un + (n+1)
    = 4 [(3/4)^n] - 1 + 4 [(3/4)] - 1




    BUT!!!!!!!! check the n+1st term... tis been so long since i've done induction that could well be wrong!!!
    from then on its clear sailing, the trick is jsut to separate Un+1 into Un + (n+1)
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    1). Prove by induction, for all positive integers n, that:

    (1 x 5) + (2 x 6) + (3 x 7) + ..... + n(n + 4) = 1/6 n(n+1)(2n+13)

    With induction you assume P(n), and prove p(1) p(n+1)

    You prove that this:

    1/6 n(n+1)(2n+13) + (n+1)((n+1) + 4)

    is equal to this:

    1/6 (n+1)((n+1)+1)(2(n+1)+13)

    Explanation:

    Your given a sequence and told the nth term is "n(n + 4)", and your told the sum of the first n terms is "1/6 n(n+1)(2n+13)". You need to prove that the sum of the first n terms + the (n+1)th term is equal to the sum of the first (n+1) termss.

    Hope that helps
 
 
 
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