​​​Prove by the principle of matchematical induction that for all n belongs to N 5^2n div by 6 gives1 as remainder whereas 5^2n-1 on div by 6 always give 5 as remainder
 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{P}\left(\mathrm{n}\right): 5^{2\mathrm{n}} \mathrm{gives}1 \mathrm{as} \mathrm{remainder} \mathrm{when} \mathrm{divided} \mathrm{by} 6 \\ \mathrm{For} \mathrm{n}=1 \\ {5}^{2\mathrm{n}}=5^2=25 \mathrm{which} \mathrm{gives} 1 \mathrm{as} \mathrm{remainder} \mathrm{when} \mathrm{divided} \mathrm{by} 6. \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{fo} \mathrm{n}=1 \\ \mathrm{Let} \mathrm{P}\left(\mathrm{n}\right) \mathrm{be} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k} \\ \Rightarrow 5^{2\mathrm{k}} \mathrm{gives}1 \mathrm{as} \mathrm{remainder} \mathrm{when} \mathrm{divided} \mathrm{by} 6. \\ \Rightarrow 5^{2\mathrm{k}}=6\mathrm{p}+1...\left(\mathrm{i}\right) \\ \mathrm{Now} \\ {5}^{2\left(\mathrm{k}+1\right)}=5^{2\mathrm{k}}.5^2 \\ =\left(6\mathrm{p}+1\right).5^2 \\ =\left(6\mathrm{p}+1\right).25 \\ =150\mathrm{p}+25 \\ =150\mathrm{p}+24+1 \\ =6\left(25\mathrm{p}+5\right)\mathbf{+}\mathbf{1} \\ \mathrm{which} \mathrm{also} \mathrm{gives} \mathrm{remainder} \mathrm{of} 1 \mathrm{when} \mathrm{divided} \mathrm{by} 6. \\ \mathrm{Hence} \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{n}=\mathrm{k}+1 \\ \mathrm{Hence} \mathrm{by} \mathrm{principal} \mathrm{of} \mathrm{Mathematical} \mathrm{Induction}, \mathrm{P}\left(\mathrm{n}\right) \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{all} \mathrm{n}\in \mathrm{N} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards