Show that $3^{4 n + 2} + 5^{2 n + 1}$ is a multiple of $14$ .

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Solution

$f (n) = 3^{4 n + 2} + 5^{2 n + 1} f (1) = 3^{4 + 2} + 5^{2 + 1} = 729 + 125 = 854 = 14 \times 61$

which is a multiple of $14$

$f (n + 1) = 3^{4 n + 6} + 5^{2 n + 3} f (n + 1) - 25 f (n) = 3^{4 n + 6} + 5^{2 n + 3} - 25 (3^{4 n + 2} - 5^{2 n + 1}) f (n + 1) - f (n) = 3^{4 n + 6} - {25.3}^{4 n + 2} + 5^{2 n + 3} - {25.5}^{2 n + 1} f (n + 1) - f (n) = 3^{4 n + 2} (81 - 25) + 5^{2 n + 3} - 5^{2 n + 3} f (n + 1) - f (n) = 56. 3^{4 n + 2}$

which is also a multiple of $14$

Hence proved.

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