Prove 2^3n – 1 is divisible by7, for all natural numbers n.
Prove 2^3n – 1 is divisible by7, for all natural numbers n.

As indicated by the inquiry,

\[P\left( n \right)\text{ }=\text{ }23n\text{ }\text{ }1\] is detachable by 7.

In this way, subbing various qualities for n, we get,

\[P\left( 0 \right)\text{ }=\text{ }20\text{ }\text{ }1\text{ }=\text{ }0\] which is separable by 7.

\[P\left( 1 \right)\text{ }=\text{ }23\text{ }\text{ }1\text{ }=\text{ }7\] which is separable by 7.

\[P\left( 2 \right)\text{ }=\text{ }26\text{ }\text{ }1\text{ }=\text{ }63\] which is detachable by 7.

\[P\left( 3 \right)\text{ }=\text{ }29\text{ }\text{ }1\text{ }=\text{ }512\] which is distinguishable by 7.

Let \[P\left( k \right)\text{ }=\text{ }23k\text{ }\text{ }1\] be distinct by 7

In this way, we get,

\[\Rightarrow 23k1\text{ }=\text{ }7x.\]

Presently, we likewise get that,

\[\Rightarrow P\left( k+1 \right)\text{ }=\text{ }23\left( k+1 \right)1\]

\[=\text{ }23\left( 7x\text{ }+\text{ }1 \right)\text{ }\text{ }1\]

\[=\text{ }56x\text{ }+\text{ }7\]

\[=\text{ }7\left( 8x\text{ }+\text{ }1 \right)\] is distinct by 7.

\[\Rightarrow P\left( k+1 \right)\] is valid when P(k) is valid.

In this way, by Mathematical Induction,

\[P\left( n \right)\text{ }=\text{ }23n\text{ }\text{ }-1\text{ }is\text{ }separable\text{ }by7,\text{ }for\text{ }all\text{ }regular\text{ }numbers\text{ }n.\]