Question

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

Answer 1

`2^3n - 7n - 1 = 8^n - 7n- 1`
`=(7+1)^n - 7n - 1`
`=C(n,0)7^n + C(n,1)7^(n-1) +... +C(n,n-2)7^2+C(n,n-1)7^1+C(n,n)7^0 - 7n - 1`
Now, `C(n,0)7^n + C(n,1)7^(n-1) +... +C(n,n-2)7^2` is clearly divisible by `49`.
So, we can write it as `49k`.
So, our expression becomes,
`= 49k +C(n,n-1)7^1+C(n,n)7^0 - 7n - 1`
`=49k+7n+1-7n-1 `
`=49k`
`:. 2^3n - 7n - 1 = 49k `
So, clearly `2^3n - 7n - 1` is divisible by `49.`