If the greatest divisor of 30 cdot 5^{2n} + 4 | vedclass

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(B) For the first expression $f(n) = 30 \cdot 5^{2n} + 4 \cdot 2^{3n}$:For $n=1$, $f(1) = 30 \cdot 25 + 4 \cdot 8 = 750 + 32 = 782 = 2 	imes 17 	ime

Vedclass · Accepted Answer

$52$

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(B) For the first expression $f(n) = 30 \cdot 5^{2n} + 4 \cdot 2^{3n}$:For $n=1$, $f(1) = 30 \cdot 25 + 4 \cdot 8 = 750 + 32 = 782 = 2 	imes 17 	imes 23$.For $n=2$, $f(2) = 30 \cdot 625 + 4 \cdot 64 = 18750 + 256 = 19006 = 2 	imes 17 	imes 559$.The greatest common divisor $p = 2 	imes 17 = 34$.For the second expression $g(n) = 2^{2n+1} - 6n - 2$:For $n=1$, $g(1) = 2^3 - 6(1) - 2 = 8 - 8 = 0$.For $n=2$, $g(2) = 2^5 - 6(2) - 2 = 32 - 14 = 18$.For $n=3$, $g(3) = 2^7 - 6(3) - 2 = 128 - 20 = 108$.The greatest common divisor $q = 	ext{gcd}(18, 108) = 18$.Thus, $p+q = 34 + 18 = 52$.

If the greatest divisor of $30 \cdot 5^{2n} + 4 \cdot 2^{3n}$ is $p, \forall n \in N$ and that of $2^{2n+1} - 6n - 2$ is $q, \forall n \in N$, then $p+q=$

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