Question

Show that ${2.7}^n+{3.5}^n-5$ is a multiple of $24$.

Answer 1

$f\left(n\right)=2.7^n+3.5^n-5f\left(1\right)=2.7^1+3.5^1-5=24$

which is a mutiple of $24$

$f(n+1)=2.7^{n+1}+3.5^{n+1}-5f(n+1)-f\left(n\right)=2.7^{n+1}+3.5^{n+1}-5-2.7^n-3.5^n+5f(n+1)-f\left(n\right)=2.7^{n+1}-2.7^n+3.5^{n+1}-3.5^{n}f(n+1)-f\left(n\right)=2.7^n(7-1)+3.5^n(5-1)f(n+1)-f\left(n\right)=12.7^n+12.5^n=12(7^n+5^n)$

which is a mutiple of $24$ as $(7^n+5^n)$ is always even

Hence proved.