$2^{3 n} - 7 n - 1$ is divisible by 49. Hence show that
$2^{3 n + 3} - 7 n - 8$ is divisible by 49, n $ϵ$ N

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Solution

We have prove that $2^{3 n}$ - 7n - 1 is divided by 49
or $(1 + 7)^{n} - n . 71$ is divided by $7^{2}$
Choosing , x = 7 in (1) We prove the result
Again $2^{3 n + 3} - 7 n - 8 = 8^{n + 1} - 7 (n + 1) - 1$
= $(1 + 7) N - 7 N - 1$ where N = n + 1
Which is divided by 49 by the case of above

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2^{3 n} - 7 n - 1

n \in N

For all natural numbers n, $2^{3 n} - 7 n - 1$ is divisible by

2^{3 n} - 7 n - 1

49

n \in N

P (n) = 2^{3 n} - 7 n - 1

P (n)

ϵ

3 \times 5^{2 n + 1} + 2^{3 n + 1}

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