Using binomial theorem, prove that `6^n-5n` always leaves remainder 1 when divided by 25.

Explanation

Using binomial expansion , we get
`(6^(n)-5n)=(1+5)^(n)-5n`
`={.^(n)C_(0)+.^(n)C_(1) xx5 +.^(n)C_(2)xx(5)^(2)+^(n)C_(3) xx(5)^(3)+^(n)C_(4) xx(5)^(4)+...+.^(n)C_(n) xx5^(n)}-5n`
`={1+5n+^nC_(2) xx(5)^(2) +^(n)C_(3) xx(5)^(3)+.^(n)C_(4) xx (5)^(4)+...+.^(n)C_(n) xx5^(n)}-5n`
`={1+.^(n)C_(2) xx (5)^(2) + .^(n)C_(3) xx(5)^(3)+.^(n)C_(4) xx(5)^(4)+...+ .^(n)C_(n) xx(5)^(n)}`
`=(5)^(2) xx{.^(n)C_(2)+ ^(n)C_(3) xx5 + ^(n)C_(4) xx(5)^(2)+...+^nC_(n) xx5^((n-2)) }+1`
` =25 xx ( " an integar ")` 1.
Hence, `(6^(n) - 5n)` when divided by 25 leaves the remainder 1.