13. Statement: 5^{2n+1}+2^{2n+1} is divisible by 7 for all n Є $\mathbb{N}$.

Proof: We can begin by proving that 5^{2n+1}+2^{2n+1} being divisible by 7 is true for the base case, n = 0.

5^{2(0)+1}+2^{2(0)+1} =

5^{1}+2^{1}=

5+2=7

7/7 =1

Based on this idea, assume the argument holds true for n=k.

5^{2k+1}+2^{2k+1}=7J where JЄ$\mathbb{N}$

2^{2k+1}=7J-5^{2k+1}

Next, we must show that the argument holds for n=k+1.

5^{2(k+1)+1}+2^{2(k+1)+1}=5^{2k+3}+2^{2k+3}

=5^{2k+1}(5^{2})+2^{2k+1}(2^{2})

=25(5^{2k+1})+4(2^{2k+1})

=25(5^{2k+1})+4(7J-5^{2k+1})

=25(5^{2k+1})+28J-4(5^{2k+1})

=21(5^{2k+1})+28J

=7[3(5^{2k+1})+4J]

7[3(5^{2k+1})+4J] will always be divisible by 7 regardless of the value of k as long as k Є $\mathbb{N}$. Hence, 5^{2n+1}+2^{2n+1} is divisible by 7 for all n Є N. ∎

Final Solution 13 Maria Acevedo(Recitation 2/21)

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