Question

Prove that ${41}^n-{14}^n$ is a multiple of 27 for all natural numbers.

Answer 1

Let the given statement be P(n), i.e.,
P(n): ${41}^n-{14}^n$ is a multiple of 27.
It can be observed that P(n) is true for n = 1 since ${41}^1-{14}^1=27$, which is a multiple of 27.
Let P (k) br true for some positive integer k, i.e.,
${41}^k-{14}^k$ is a multiple of 27
$\therefore {41}^k-{14}^k=27m,wheremϵN$ . . . . .(1)
We shall now prove that P(k+1) is true whenever P(k) is true.
Consider
${41}^{k+1}-{14}^{k+1}$
$={41}^k.41-{14}^k.14$
$=41({41}^k-{14}^k+{14}^k)-{14}^k.14$
$=41({41}^k-{14}^k)+{41.14}^k-{14}^k.14$
$=41.27m+{14}^k(41-14)$
$=41.27m+{27.14}^k$
$27(41m-{14}^k)$
$=27\times r,wherer=(41m-{14}^k)$ is a natural number
Therefore, ${41}^{k+1}-{14}^{k+1}$ is a multiple of 27 .
Thus, P(k+1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.