Question

Prove by the Principle of Mathematical Induction: For any natural number n, $7^n-2^n$ is divisible by 5.

Answer 1

Assume given statement

Let $P\left(n\right)=7^n-2^n$ is divisible by $5\forall nϵN$.

Check that statement is true for $n=1$

$\because P\left(1\right)=7^1-2^1=7-2=5$

$\Rightarrow P\left(1\right)$is divisible by 5

So, P(n) is true for $n=1.$

Let assume P(k) is true $\forall nϵN$

Assume P(k) to be true and then prove $P(k+1)$ is true.

So, P(k) is divisible by 5

$Rightarrow{7}^k-2^k=5q,$where $qϵN$…(1)

$P(k+1)=7^k+1-2^k+1$

$={7.7}^k-{2.2}^k$

$=(5+2){.7}^k-{2.2}^k$

$={5.7}^k+{2.7}^k-{2.2}^k$

$={5.7}^k+2(7^k-2^k)P(k+1)$

$={5.7}^k+2\left(5q\right)=5(7^k+2q),$ where $qϵN$

$\therefore P(k+1)$ divisible by 5

Hence, $P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.