Question

Show that $9^{n+1}-8n-9$ is divisible by $64$, whenever $n$ is a positive integer

Answer 1

Consider,
$9^{n+1}=(1+8{)}^{n+1}{=}^{n+1}{C}_0{+}^{n+1}{C}_1(8\left){+}^{n+1}{C}_2\right(8{)}^2+.....{+}^{n+1}{C}_{n+1}(8{)}^{n+1}$

$=1+(n+1)\left(8\right)+8^2{[}^{n+1}{C}_2{+}^{n+1}{C}_{3}x8+.....{+}^{n+1}{C}_{n+1}\left(8{)}^{n-1}\right]$

$=9+8n+64{[}^{n+1}{C}_2{+}^{n+1}{C}_{3}x8+....{+}^{n+1}{C}_{n+1}\left(8{)}^{n-1}\right]$

$\Rightarrow 9^{n+1}-8n-9=64k,$

Where $k{=}^{n+1}{C}_2{+}^{n+1}{C}_{3}x8+....{+}^{n+1}{C}_{n+1}(8{)}^{n-1}$ which is a natural number

Thus, $9^{n+1}-8n-9$ is divisible by $64$, whenever $n$ is a positive integer.