Using binomial theorem, prove that `8^(n)-7n` always leaves remainder 1 when divided by 49.

Explanation

For any two numbers a and b if we can find the numbers q and r such that `a=bq+r`, then we say that b divides a with q as quotient and r as remaindder. Thus, in order to show that `8^(n)-7n` leaves remainer 1 when divided by 49, we prove that `8^(n)-7n=49k+1`, where k is some natural number.
Now,
`8^(n)-7` can be written as `(1+7)^(n)-7n`
Now, we can write
`(1+7)^(n)+{1+^(n)C_(1)7^(1)+.^(n)C_(2)7^(2)+.^(n)C_(3)7^(3)+ . . .+.^(n)C_(n)7^(n)}-7n`
`={1+7n+7^(2).^(n)C_(2)+7^(3).^(n)C_(3)+. . . .+7^(n)}-7n`
`=1+7^(2)({.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)}`
`=1+49{.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)}`
`therefore8^(n)-7n=49k+1,` where `K=.^(n)C_(2)+7.^(n)C_(3)+ . . .+7^(n-2)`
this shows that when `8^(n)-7n` is divided by 49 always leaves remainder as 1.