Question

Prove that ${2.7}^n+{3.5}^n-5$ is divisible by 24 true for all natural numbers.

Answer 1

Let p (n) be the statement given by
P(n): ${2.7}^n+{3.5}^n-5$ is divisible by 24
Then p(1): $2\times 7^1+3\times 5^1-5=14+15-5=24$ which is divisible by 24
$\therefore$ p(1) is true
Let p(m): ${2.7}^m+{3.5}^m-5$ is divisible by 24 be true
$\Rightarrow {2.7}^m+{3.5}^m-5=24\lambda ;\lambda ϵN$
$\Rightarrow {3.5}^m=24\lambda +5-{2.7}^m$ . . . . . (1)
Now
$p(m+1):{2.7}^{m+1}+{3.5}^{m+1}-5$
$={2.7}^{m+1}+\left({3.5}^m\right)5-5$
$={2.7}^{m+1}+(24\lambda +5-{2.7}^m)5-5\left(from\right(1\left)\right)$
$={2.7}^{m+1}+120\lambda +25-{10.7}^m-5$
$=({2.7}^{m+1}-{10.7}^m)+120\lambda +20$
$=(2\times 7\lambda 7^m-10\times 7^m)+120\lambda +24-4$
$=(14-10)7^m-4+24(5\lambda +1)$
$=4(7^m-1)+24(5\lambda +1)$
$=4\times 6\mu +24(5\lambda +1)(\because 7^m-1$ is a multiple of 6 for all $mϵN\therefore 7^m-1=6\mu ,\mu ϵN$)
$=24(\mu +5\lambda +1)$ which is divisible by 24
$\therefore$ p(m+1) is true.
Hence by principle of mathematical induction p(n) is true for all $nϵN$