If 2 cdot 4^{2n+1} + 3^{3n+1} is divisible by | vedclass

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(B) Let $P(n) = 2 \cdot 4^{2n+1} + 3^{3n+1} = 2 \cdot (2^2)^{2n+1} + 3^{3n+1} = 2 \cdot 2^{4n+2} + 3^{3n+1} = 2^{4n+3} + 3^{3n+1}$.For $n=1$,$P(1) = 2

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(B) Let $P(n) = 2 \cdot 4^{2n+1} + 3^{3n+1} = 2 \cdot (2^2)^{2n+1} + 3^{3n+1} = 2 \cdot 2^{4n+2} + 3^{3n+1} = 2^{4n+3} + 3^{3n+1}$.For $n=1$,$P(1) = 2^{4(1)+3} + 3^{3(1)+1} = 2^7 + 3^4 = 128 + 81 = 209$.For $n=2$,$P(2) = 2^{4(2)+3} + 3^{3(2)+1} = 2^{11} + 3^7 = 2048 + 2187 = 4235$.We find the greatest common divisor of $209$ and $4235$.$209 = 11 	imes 19$.$4235 = 11 	imes 385$.The greatest common divisor is $11$.Thus,$P(n)$ is divisible by $11$ for all $n \in N$.

If $2 \cdot 4^{2n+1} + 3^{3n+1}$ is divisible by $k$ for all $n \in N$,then $k=$

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