

InterviewSolution
Saved Bookmarks
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.` |
|