Question

For each $nϵN$, then $3^{2n+1}+1$ is divisible by -

• A. $2$
• B. $3$
• C. $7$
• D. None of these

Answer 1

Answer: (A)

$2$

$3^{2n+1}+1$

Let $n=1$.

Then, $3^{2n+1}+1$

$=3^{(2\times 1)+1}+1$

$=3^{2+1}+1$

$3^3+1=28$ which is divisible by $2.$

Let $n=2$.

Then, $3^{2n+1}+1$

$3^5+1$

$243+1=244$ which is divisible by $2.$