For any natural number n,(15 times 5^{2n}) + | vedclass

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(D) Let $P(n) = (15 	imes 5^{2n}) + (2 	imes 2^{3n})$.For $n = 1$,we have:$P(1) = (15 	imes 5^2) + (2 	imes 2^3) = (15 	imes 25) + (2 	imes 8) =

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$17$

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(D) Let $P(n) = (15 	imes 5^{2n}) + (2 	imes 2^{3n})$.For $n = 1$,we have:$P(1) = (15 	imes 5^2) + (2 	imes 2^3) = (15 	imes 25) + (2 	imes 8) = 375 + 16 = 391$.Now,check the divisibility of $391$ by the given options:$391 / 7 \approx 55.85$$391 / 11 \approx 35.54$$391 / 13 \approx 30.07$$391 / 17 = 23$.Since $391$ is divisible by $17$,the expression $(15 	imes 5^{2n}) + (2 	imes 2^{3n})$ is divisible by $17$ for all $n \in N$.

For any natural number $n$,$(15 \times 5^{2n}) + (2 \times 2^{3n})$ is divisible by

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