If 2(4^{2n+1}) + 3^{3n+1} is divisible by k,w | vedclass

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(C) Let $P(n) = 2(4^{2n+1}) + 3^{3n+1}$.For $n = 1$,$P(1) = 2(4^3) + 3^4 = 2(64) + 81 = 128 + 81 = 209$.We know that $209 = 11 	imes 19$.For $n = 2$,

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$11$

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(C) Let $P(n) = 2(4^{2n+1}) + 3^{3n+1}$.For $n = 1$,$P(1) = 2(4^3) + 3^4 = 2(64) + 81 = 128 + 81 = 209$.We know that $209 = 11 	imes 19$.For $n = 2$,$P(2) = 2(4^5) + 3^7 = 2(1024) + 2187 = 2048 + 2187 = 4235$.Checking divisibility for $P(2)$:$4235 / 11 = 385$ (divisible).$4235 / 19 = 222.89$ (not divisible).Since the expression must be divisible by $k$ for all $n \in N$,we test $k = 11$.$P(n) = 2 \cdot 4 \cdot 16^n + 3 \cdot 27^n = 8(16^n) + 3(27^n)$.Modulo $11$:$16 \equiv 5 \pmod{11}$ and $27 \equiv 5 \pmod{11}$.$P(n) \equiv 8(5^n) + 3(5^n) = 11(5^n) \equiv 0 \pmod{11}$.Thus,the expression is always divisible by $11$.

If $2(4^{2n+1}) + 3^{3n+1}$ is divisible by $k$,where $k > 1$,for all $n \in N$,then the value of $k$ is:

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