For all n in N,2^{2n+1} + 3^{2n+1} is divisib | vedclass

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(B) Let $P(n)$ be the statement $2^{2n+1} + 3^{2n+1}$.For $n = 1$:$P(1) = 2^{2(1)+1} + 3^{2(1)+1} = 2^3 + 3^3 = 8 + 27 = 35$.Since $35$ is divisible b

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$5$

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(B) Let $P(n)$ be the statement $2^{2n+1} + 3^{2n+1}$.For $n = 1$:$P(1) = 2^{2(1)+1} + 3^{2(1)+1} = 2^3 + 3^3 = 8 + 27 = 35$.Since $35$ is divisible by $5$,$P(1)$ is true.Assume $P(m)$ is true for some $m \in N$,i.e.,$2^{2m+1} + 3^{2m+1} = 5k$ for some integer $k$.Then $2^{2m+1} = 5k - 3^{2m+1} \quad (i)$.Now,consider $P(m+1)$:$P(m+1) = 2^{2(m+1)+1} + 3^{2(m+1)+1} = 2^{2m+3} + 3^{2m+3}$.$P(m+1) = 2^2 \cdot 2^{2m+1} + 3^2 \cdot 3^{2m+1}$.Substituting from $(i)$:$P(m+1) = 4(5k - 3^{2m+1}) + 9 \cdot 3^{2m+1}$.$P(m+1) = 20k - 4 \cdot 3^{2m+1} + 9 \cdot 3^{2m+1}$.$P(m+1) = 20k + 5 \cdot 3^{2m+1} = 5(4k + 3^{2m+1})$.Since $5(4k + 3^{2m+1})$ is divisible by $5$,$P(m+1)$ is true.By the principle of mathematical induction,$2^{2n+1} + 3^{2n+1}$ is divisible by $5$ for all $n \in N$.

For all $n \in N$,$2^{2n+1} + 3^{2n+1}$ is divisible by

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