Question

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

Answer 1

We have,

$2^{4n+4}-15n-16=2^{4(n+1)}-15n-16$

$={16}^{n+1}-15n-16=(1+15{)}^{n+1}-15n-16$

${=}^{n+1}C-0{15}^0{+}^{n+1}{C}_1{15}^1{+}^{n+1}{C}_2{15}^2{+}^{n+1}{C}_3{15}^3+....{+}^{n+1}{C}_{n+1}(15{)}^{n+1}-15n-16$

$=1+(n+1)15{+}^{n+1}{C}_2{15}^2{+}^{n+1}{C}_3{15}^3+....{+}^{n+1}{C}_{n+1}(15{)}^{n+1}-15n-16$

$=1+15n+15{+}^{n+1}{C}_2{15}^2{+}^{n+1}{C}_3{15}^3+...{+}^{n+1}{C}_{n+1}(15{)}^{n+1}-15n-16$

$={15}^2{[}^{n+1}{C}_2{+}^{n+1}{C}_{3}15+....soon]$

Thus, $2^{4n+4}-15n-16$ is divisible by 225.