PROVE BY M.I (41)n-(14) is multiple of 27
The question is wrong its 41n-41n
p(k)=41k-14k
=27m
41k=27m+14k
p(k+1)=41k+1-14k+1
=41k.41-14k.14
=(27m+14k)41-14k.14
=(27m)41+(14k)41-14k.14
=(27m)41+14k(41-14)
=(27m)41+14k(27)
=27(41m+14k)
=27q, where q=(41m+14k)