meredith48034
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Assume that the following statement, S_k, is true:
A factor of (2^(2k-1) + 3^(2k-1)))is 5.
Show that for each integer k ≥ 1, if S_k is true, then S_(k+1) is true.
2^2(k + 1) - 1)) + 3^2(k + 1) - 1)) = 2^(2k + ____) - 1)) + 3^2k + ___) - 1))
= 2^(2k - 1) * 2^(___) + 3^(2k - 1) * 3^(___)
= ______ * 2^(2k-1) + ______ * 3^(2k-1)
= (2^(2k-1) + 3^(2k-1)) + (2^(2k-1) + 3^(2k-1)) + (2^(2k-1) + (3^(2k-1) + ______*3^(2k-1)

Hence, S_k+1 is true or false, which completes the inductive step and the proof by the mathematical induction.



Answer :

mhanifa

It is assumed the statement is true for [tex]S_k[/tex]:

  • A factor of [tex]2^{2k-1}+3^{2k-1}[/tex] is 5.

Show that [tex]S_{k+1}[/tex] is also divisible by 5:

  • [tex]2^{2(k+1)-1}+3^{2(k+1)-1}=[/tex]
  • [tex]2^{2k+2-1}+3^{2k+2-1}=[/tex]
  • [tex]2^{(2k-1)+2}+3^{(2k-1)+2}=[/tex]
  • [tex]2^{2k-1}*2^2+3^{2k-1}*3^2=[/tex]
  • [tex]4*2^{2k-1}+9*3^{2k-1}=[/tex]
  • [tex]4*2^{2k-1}+4*3^{2k-1}+5*3^{2k-1}=[/tex]
  • [tex]4(2^{2k-1}+3^{2k-1})+5*3^{2k-1}[/tex]

The first part is divisible by 5 as we assumed in the beginning and the second part has a factor of 5 so the sum is divisible by 5.

Hence it makes [tex]S_{k+1}[/tex] true, so the proof is complete.

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