Question

Show that $2^{2n}-3n-1$ is divisible by $9$ for all positive integral values of $n$.

Answer 1

Let the statement be $2^{2n}-3n-1$ is divisble by $9$

Let us check for $n=1$

$\Rightarrow$ $2^{2\times 1}-3\times 1-1=4-3-1=0$

$0$ is divisible by $9$.

Assume the statement is true for $n=k$

$2^{2k}-3k-1=9J$ $J\in N$

$2^{2k}=9J+3k+1$ .....(i)

Let us check whether the statement is true for $n=k+1$

$=2^{2(k+1)}-3(k+1)-1$

$=2^{2k}\times 2^2-3k-3-1$

$=4(9J+3k+1)-3k-4$ ....From eq (i)

$=36J+12k+4-3k-4$

$=9(4J+k)$ [ Here $(4J+k)$ will always an integer ]

Hence, $2^{2n}-3n-1$ is always divisible by $9$.