Question

$\forall n\in N;7^{4n+3}-18$ is divisible by

• A. $5$
• B. $7$
• C. $25$
• D. $40$

Answer 1

Answer: (A), (C)

$5$
$25$

$P\left(n\right)=7^{4n+3}-18=343{(50-1)}^{2n}-18$
$\Rightarrow P\left(n\right)=343({50}^{2n}-{2n}_{C_1}{50}^{2n-1}+{2n}_{C_2}{50}^{2n-2}-.....{-2n}_{C_{2n-1}}50+{2n}_{C_{2n}})-18$
$\Rightarrow P\left(n\right)=50({50}^{2n-1}-{2n}_{C_1}{50}^{2n-2}+{2n}_{C_2}{50}^{2n-3}-.....{-2n}_{C_{2n-1}})+343-18$
$\Rightarrow P\left(n\right)=50\lambda +325=25k$
Therefore, $P\left(n\right)$ is divisible by $5$ and $25$.
Ans: A,C