Using binomial theorem, prove that `2^(3n)-7^n-1` is divisible by `49` , where `n in Ndot`

Explanation

`2^(3n) - 7n - 1= (2^(3))^(n) - 7n - 1`
` = (1+7)^(n) - 7n - 1`
`= (1+7)^(n)-7n-1`
`= 1 + 7n + .^(n)C_(2)7^(2)+ .^(n)C_(3)7^(3)+"…….".^(n)C_(n)7^(n) - 7n - 1`
` = 7^(2)[.^(n)C_(2)+.^(n)C_(3)7+"……"+.^(n)C_(n)7^(n_2)] = 49K(1)`
where K is an integer.
Therefore, `2^(3n)-7n-1` is divisible by 49. Now,
`2^(3n+3) - 7n - 8 = 2^(3)2^(3n)-7n-8`
` = 8(2^(3n)-7n-1)+49`
` = 8 xx 49K + 49n` [From (1)]
`= 49 (8K + n)`
Therefore, `2^(3n+3)-7n-8` is divisible by 49.