Prove this :- 41^n - 14^n is a multiple of 27 ?
Prove this :- 41^n - 14^n is a multiple of 27 ?
Any number which is a multiple of 27 is in the form of-->
any natural number ×27,
let p(n):41^n-14^n=27d,where d belongs to natural number,
for n=1,
L.H.S=41¹-14¹=41-14=27=27×1=R.H.S,
hence L.H.S=R.H.S so, p(n) is true for n=1,
assume that p(k) is true,
p(k)=41^k-14^k=27 m ,where m belongs to natural number,----(1).
we will prove that p(k+1) is true,
p(k+1),
LHS= 41^(k+1)-14^(k+1),
=41^k.41^1-14^k.14^1,
from (1).41^k=27m+14^k,
=41(27m+14^k)-14^k.14^1,
=41.27m+41.14^k-14^k.14^1,
=41.27m+14^k(41-14),
=41.27m +14^k(27),
=(27)(27m+14^k)=27r,
where r =27m+14^k,
so, p(k+1) is true whenever p(k) is true,
so, by the principle of Mathematical Induction p(n) is true for 41^n-14^n is a multiple of 27.
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