Among the statements : (S1) : 2023^2022-1999^2022 is divisible by 8 (S2) : 13(13)^n-12 n-1…

Question

Among the statements :
(S1) : $2023^{2022}-1999^{2022}$ is divisible by 8
(S2) : $13(13)^{n}-12 n-13$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$

Accepted Answer

We have, ${S}_{1}$ : $(2023{)}^{2022}-(1999{)}^{2022}$

$\begin{gathered} =(1999+24{)}^{2022}-(1999{)}^{2022}={}^{2022}{C}_{0}(1999{)}^{2022}(24{)}^{0} \ +{}^{2022}{C}_{1}(1999{)}^{2021}(24{)}^{1}+{}^{2022}{C}_{2}(1999{)}^{2020}(24{)}^{2} \ +\ldots -(1999{)}^{2022} \ ={}^{2022}{C}_{1}(1999{)}^{2021}(24)+{}^{2022}{C}_{2}(1999{)}^{2022}(24{)}^{2} \ =24({}^{2022}{C}_{1}(1999{)}^{2021}+{}^{2022}{C}_{2}(1999{)}^{2022}(24)+\ldots +\ldots ) \ \Rightarrow {S}_{1}24 \end{gathered}$

$\begin{gathered} {S}_{2}:13(13{)}^{n}-12n-13 \ \ {13}^{n}=(1+12{)}^{n} \ \ \, =1+{}^{n}{C}_{1}12+{}^{n}{C}_{2}(12{)}^{2}+{}^{n}{C}_{3}(12{)}^{3} \ \ 	herefore {S}_{2}: 13(1+{}^{n}{C}_{1}(12)+{}^{n}{C}_{2}(12{)}^{2}+{}^{n}{C}_{3}(12{)}^{3}+\ldots )-12n-13 \ \ \, = 13+156n+13({}^{n}{C}_{2}(12{)}^{2}+{}^{n}{C}_{3}(12{)}^{3}+\ldots )-12n-13 \ \ \, = 144	imes 13({}^{n}{C}_{2}+{}^{n}{C}_{3}(12)+\ldots ) \end{gathered}$

$\Rightarrow {S}_{2}$ is divisible by 144 for infinitely many $n\in N$

Solutions · 1