Question

Prove ${41}^n-{14}^n\text{is a multiple of 27.}$

Answer 1

$\text{Let}P\left(n\right)={41}^n-{14}^n\text{is a multiple of 27}\text{for n = 1}P\left(1\right)={41}^1-{14}^1\text{is a multiple of 27}\Rightarrow 27\text{is a multiple of 27}\therefore P\left(1\right)\text{is true}\text{Let P(n) be true for n = k}\therefore P\left(k\right)={41}^k-{14}^k\text{is a multiple of 27}={41}^k-{14}^k=27\lambda \text{For}n=k+1P(k+1)={41}^{k+1}-{14}^{k+1}\text{is a multiple of 27}\text{Now}{41}^{k+1}-{14}^{k+1}={41}^{k+1}-{41}^k.14+{41}^k.14-{14}^{k+1}={41}^k(41-14)+14({14}^k-{14}^k)={41}^k\times 27+14\times 27\lambda =27({41}^k+14\lambda )\Rightarrow {41}^{k+1}-{14}^{k+1}\text{is a multiple of 27}\therefore P(k+1)\text{is true}\text{thus P(k) is true}\Rightarrow P(k+1)\text{is true}$