Question

$3^{2n-2}-8n-9$ is divisible by $64$

Answer 1

$3^{2n-2}-8n-9=3^2{.3}^{2n}-8n-9$

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

$=9[1{+}^n{C}_{1}8{+}^n{C}_2{8}^2{+}^n{C}_3{8}^3+.....{.}^n{C}_{n}8]-8n-9$

$=9{+}^n{C}_1.72+9{[}^n{C}_2{8}^2+....{.}^n{C}_n{8}^4]-8n-9$

$=72n-8n+9{[}^n{C}_2{8}^2{+}^n{C}_3{8}^3+...{.}^n{C}_n{8}^n]$

$=64n+9\times 64{[}^n{C}_2{+}^n{C}_3.8+....{.8}^{n-2}]$

$=64[n+9{(}^n{C}_2{+}^n{C}_3.8+....{.8}^{n-2})]$

$=64\times k.....$ (where k is an integer)

Hence,

$(3^{2n-2}-8n-9)$ is divisible by $64$.