If n is a positive integer,then 2 cdot 4^{2n+ | vedclass

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(C) Let $p(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}$.For $n = 1$,$p(1) = 2 \cdot 4^{2(1)+1} + 3^{3(1)+1} = 2 \cdot 4^3 + 3^4 = 2 \cdot 64 + 81 = 128 + 81 = 20

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Let $p(n) = 2 \cdot 4^{2n+1} + 3^{3n+1}$.For $n = 1$,$p(1) = 2 \cdot 4^{2(1)+1} + 3^{3(1)+1} = 2 \cdot 4^3 + 3^4 = 2 \cdot 64 + 81 = 128 + 81 = 209$.Since $209 = 11 	imes 19$,$p(1)$ is divisible by $11$.Assume $p(k) = 2 \cdot 4^{2k+1} + 3^{3k+1}$ is divisible by $11$ for some positive integer $k$.Now,consider $p(k+1) = 2 \cdot 4^{2(k+1)+1} + 3^{3(k+1)+1} = 2 \cdot 4^{2k+3} + 3^{3k+4}$.$p(k+1) = 2 \cdot 4^{2k+1} \cdot 4^2 + 3^{3k+1} \cdot 3^3 = 16 \cdot (2 \cdot 4^{2k+1}) + 27 \cdot 3^{3k+1}$.$p(k+1) = 16 \cdot (2 \cdot 4^{2k+1} + 3^{3k+1}) + (27 - 16) \cdot 3^{3k+1}$.$p(k+1) = 16 \cdot p(k) + 11 \cdot 3^{3k+1}$.Since $p(k)$ is divisible by $11$ and $11 \cdot 3^{3k+1}$ is clearly divisible by $11$,$p(k+1)$ is divisible by $11$.By the principle of mathematical induction,$2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $11$ for all positive integers $n$.

If $n$ is a positive integer,then $2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by

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