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Re: [PrimeNumbers] solve using induction to find the sequence

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  • mikeoakes2@aol.com
    In a message dated 29/04/03 22:51:04 GMT Daylight Time, luken2u@aol.com ... Assume that, for 0
    Message 1 of 2 , Apr 29 3:08 PM
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      In a message dated 29/04/03 22:51:04 GMT Daylight Time, luken2u@...
      writes:


      > i have a problem that i can't figure out.. and i need to know how to
      > do this b4 10 tonight... here is the problem (using induction)
      >
      > prove by induction
      > let e (sub 0) = 1,
      > e (sub 1) = 4
      > and for n >1.
      > let e sub n = 4[e (sub n - 1) - e (sub n-2)]
      >
      > what are the first six terms of the sequence?
      >
      > prove that e(sub n) = 2^n(n+1)
      >

      Assume that, for 0 <= k < n
      e_k = (k+1)*2^k (1)

      Then
      e_n = 4*[e_(n-1) - e(n-2)]
      = 4*[n*2^(n-1) - (n-1)*2^(n-2)]
      = 2*n*2^n - (n-1)*2^n
      = (n+1)*2^n
      So if (1) is true for k < n, it is true also for k = n.
      But (1) is true for n = 0 and n = 1.
      So (1) is true for all n >= 0.
      Q.E.D.

      Mike Oakes


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