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  • 23n -1 is divisible by7, for all natural numbers n.

2^{3n} -1 is divisible by7, for all natural numbers n.

 

Answers (1)

Now, we'll substitute different values for n,

 \\ \mathrm{P}(\mathrm{0})=2^{0}-1=0,\text{ divisible by 7} \\ P(1)=2^{3}-1=7,\text{ divisible by 7}\\ P(2)=2^{6}-1=63,$ divisible by 7\\ $P(3)=2^{9}-1=512, divisible by 7\\

 Now, let us consider, 

\\ P(k)=2^{3 k}-1\ be \ divisible \ by \ 7 \\\ Thus, \ 2^{2 k}-1=7 x\\

We also get that,

P(k+1)=2^{3(k+1)}-1=2^{3}(3 x+1)-1=56 x+7, is\ divisible \ by\ 7

Thus, \mathrm{P}(\mathrm{k}+1)\ is\: \: also\: \: true,\\

Hence, by mathematical induction, For each natural no. n it is true that, 

\mathrm{P}(\mathrm{n})=2^{3 \mathrm{n}}-1 \ is\ divisible \ by \ 7 .

 

Posted by

infoexpert21

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