Question

Use mathematical induction to prove that ${2.7}^n+{3.5}^n-5$ is divisible by 24 for all n > 0.

Answer 1

Let $A_n={2.7}^n+{3.5}^n-5$

Then, $A_1=2.7+3.5-5=14+15-5=24$

Hence $A_1$ is divisible by 24.

Now assume that $A_m$ is divisible by 24 so that we may write $A_m={2.7}^m+{3.5}^m-5=24KK\in N$............(1)

$A_{m+1}-A_m=2(7^{m+1}-7^m)+3.\left(5^{m+1-5^m}\right)-5+5$

$={2.7}^m(7-1)+{3.5}^m(5-1)=12(7^m+5^m)$

Since $7^m$ and $5^m$ are odd integers for all $$m \, \in \, N$,

their sum must be an even integer , Say

$7^m+5^m=2p,,p\in N$

Hence, $A_{m+1}-A_m=12.2p=24p$

or $A_{m+1}=A_m+24p=24K+24p,$ by (1)

Hence $A_{m+1}$ is divisible by 24.

If follows by mathematical induction that $A_n$ is divisible by 24 for all $n\in N$