`4 1^n-1 4^n` is a multiple of `27 `

Explanation

`" let "P(n) =41^(n)-14^(n)`
For n=1
`P(1) =41^(1)-14^(1) =41 -14 = 27 =27 xx 1`
Which is divisible by 27
`:. P (n)` is true for n=1
Let P (n) be true for n=K
`:.P(k) =41^(k)-14^(k) =27lambda ("say ")`
`" Where " lambda in I`
For n=K+1
`P(k+1)=41^(k+1)-14^(k+1)=41.41^(k)-14.14^(k)`
`=41.41^(k) -14.41^(k)+14.41^(k)-14.14^(k)`
`=41^(k)(41-14)+14(41^(k)-14^(k))`
`=27.41^(k)+14.27 lambda(41^(k)+14lambda)`
[From equation (1)]
Which is divisible by 27.
`rArr P (n) ` is also true for n=k=1
Hence from the principle of mathematical induction P(n) is true for all natural numbers n.