Question

Show that $9^{n+1}-8n-9$ is $÷$ by $64$ for $+$ ve integer $n$ , using binomial theorem.

Answer 1

To prove

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

$(9{)}^{n+1}=(1+8{)}^{n+1}$

${=}^{n+1}{\text{C}}_0{+}^{n+1}{\text{C}}_{1}8+........{+}^{n+1}{\text{C}}_{n+1}(8{)}^{n+1}$

$=1+\frac{(n+1)n!}{n!}\left(8\right)+.......+(8{)}^{n+1}$

$=8n+9{+}^{n+1}{\text{C}}_2(8{)}^2+.....+(8{)}^{n+1}$

$\therefore 9^{n+1}-8n-9{=}^{n+1}{\text{C}}_2(8{)}^2+.....+(8{)}^{n+1}$

$=(8{)}^2{(}^{n+1}{\text{C}}_2+.....+(8{)}^{n+1-2})$

$9^{n+1}-8n-9=(8{)}^{2}k$

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