1.

Show that `2^(4n+4)-15n-16`, where `n in N`is divisible by `225.`

Answer» `2^(4n+4) - 15n - 16 = 16^(n+1) - 15n - 16`
`=(15+1)^(n+1) - 15n - 16`
`=C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2+C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16`
Now, `C(n+1,0)15^(n+1) + C(n+1,1)15^(n) +... +C(n+1,n-1)15^2` is clearly divisible by `225`.
So, we can write it as `225k`.
So, our expression becomes,
`= 225k +C(n+1,n)15^1+C(n+1,n+1)15^0 - 15n - 16`
`=225k+15(n+1)+1-15n-16 `
`=225k+15n+15+1-15n-16`
`=225k`
`:. 2^(4n+4) - 15n - 16 = 225k `
So, clearly `2^(4n+4) - 15n - 16` is divisible by `225.`


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