Question

Using binomial theorem, prove that $3^{3n+2}-8b-9$ is divisible by $64,n\in N.$

Answer 1

$3^{3n+2}-8b-9$

$=3^{3n+1}-8b-9=3^{n+1}-8n-9$

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

$={(}^{n+1}{C}_0{+}^{n+1}{C}_1{8}^1{+}^{n+1}{C}_2{8}^2+.....{+}^{n+1}{C}_{n+1}{8}^{n+1})-8n-9$

$=(1+8(n+1)+{64}^{n+1}{C}_2+...+64(8{)}^{n-1})-8n-9$

$=64{(}^{n+1}{C}_2+....+8^{n-1})$

Thus, $3^{2n+2}-8n-9$ is divisible by 64.