Question

Using binomial theorem, show that $9^n-8n-1$ is always divisible by $64$.

Answer 1

$9^n-8n-1=(1+8{)}^n-8n-1=1{+}^n{C}_1(8\left){+}^n{C}_2\right(8{)}^2+....{+}^n{C}_n(8{)}^n-8n-1$

${=}^n{C}_2\left(8{)}^2{+}^n{C}_3\right(8{)}^3+.....{+}^n{C}_n(8{)}^n=8^{2}k=64k$

$⟹9^n-8n-1$ is always divisible by $64$