Prove by PMI :2^2n - 1 is divisible by 3 and 2n+1<2^n , for all natural numbers n ≥ 3.
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p(n) = 2^(2n) - 1
For n=0 is true, since
p(0)=2^(0) - 1
p(0) =0
Suppose is true for some n = k ≥ 1, that is
p(k) = 2^(2k) - 1 = 3R ------(1) {R is includin z+}
Prove that (1.2) is true for n = k + 1, that is
n=k+1;
p(k+1) = 2^(2k+1) - 1
=2^(2k)*4 - 1-3 + 3
=2(2^(2k) - 1) + 3
=2(3R)+ 3
= 3{2R + 1 }
For n=0 is true, since
p(0)=2^(0) - 1
p(0) =0
Suppose is true for some n = k ≥ 1, that is
p(k) = 2^(2k) - 1 = 3R ------(1) {R is includin z+}
Prove that (1.2) is true for n = k + 1, that is
n=k+1;
p(k+1) = 2^(2k+1) - 1
=2^(2k)*4 - 1-3 + 3
=2(2^(2k) - 1) + 3
=2(3R)+ 3
= 3{2R + 1 }
thereby showing that indeed P(k + 1) holds.
Since both the basis and the inductive step have been performed, by mathematical induction, the statement P(n) holds for all natural numbers n.
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