Question

Show that $2^{4n+4}-15n-16,$ where $nϵN$ is divisible by 225.

Answer 1

We have
$2^{4n+4}-15n-16=2^{4(n+1)}-15n-16={16}^{(n+1)}-15n-16.\text{Now},{16}^{n+1}-15n-16=(1+15{)}^{n+1}-15n-16{=}^{n+1}{C}_0{+}^{n+1}{C}_1\times 15{+}^{n+1}{C}_2\times (15{)}^2{+}^{n+1}{C}_3\times (15{)}^3+\cdots {+}^{n+1}{C}_{n+1}\times (15{)}^{n+1}-15n-16=1+(n+1)\times 15{+}^{n+1}{C}_2(15{)}^2{+}^{n+1}{C}_3\times (15{)}^3+\cdots +(15{)}^{n+1}-15n-16{=}^{n+1}{C}_2\times (15{)}^2{+}^{n+1}{C}_3\times (15{)}^3+\cdots +(15{)}^{n+1}=(15{)}^2\times {[}^{n+1}{C}_2{+}^{n+1}{C}_3\times 15+\cdots +(15{)}^{n-1}]=225\times \text{(an integer)}$