Question

Let $f\left(n\right)={10}^n+3\cdot 4^{n+2}+5;\forall n\in N$. The greatest value of the integer which divides $f\left(n\right);\forall n$, is

• A. $27$
• B. $3$
• C. $9$
• D. none of these

Answer 1

The correct option is C $9$

$P\left(n\right)={10}^n+3.4^{n+2}+5={(1+9)}^n+48{(1+3)}^n+5$
$\Rightarrow P\left(n\right)=(1+n_{C_1}9+n_{C_2}{9}^2+...+n_{C_n}{9}^n)+48(1+n_{C_1}3+n_{C_2}{3}^2+...+n_{C_n}{3}^n)+5$
$\Rightarrow P\left(n\right)=54+9(n_{C_1}+n_{C_2}9+...+n_{C_n}{9}^{n-1})+16.9(n_{C_1}+n_{C_2}3+...+n_{C_n}{3}^{n-1})$
Therefore, $P\left(n\right)$ is divisible by $9$.
Ans: C