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Home » Prove For any natural number n, 7^n – 2^n is divisible by 5.
Prove For any natural number n, 7^n – 2^n is divisible by 5.
CBSE | Class 11 | NCERT Exemplar | Principle of Mathematical Induction
Prove For any natural number n, 7^n – 2^n is divisible by 5.

As indicated by the inquiry,

    \[~P\left( n \right)\text{ }=\text{ }7n\text{ }\text{ }2n\]

is distinct by 5.

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

    \[~P\left( 0 \right)\text{ }=\text{ }70\text{ }\text{ }20\text{ }=\text{ }0\]

Which is distinguishable by 5.

    \[~P\left( 1 \right)\text{ }=\text{ }71\text{ }\text{ }21\text{ }=\text{ }5\]

Which is distinguishable by 5.

    \[~P\left( 2 \right)\text{ }=\text{ }72\text{ }\text{ }22\text{ }=\text{ }45\]

Which is distinguishable by 5.

    \[~P\left( 3 \right)\text{ }=\text{ }73\text{ }\text{ }23\text{ }=\text{ }335\]

Which is distinguishable by 5.

Let

    \[P\left( k \right)\text{ }=\text{ }7k\text{ }\text{ }2k\]

be distinguishable by 5

Thus, we get,

    \[\begin{array}{*{35}{l}} ~ \\ \Rightarrow 7k2k\text{ }=\text{ }5x. \\ \end{array}\]

Presently, we additionally get that,

    \[\begin{array}{*{35}{l}} ~ \\ \Rightarrow P\left( k+1 \right)=\text{ }7k+12k+1 \\ \end{array}\]

    \[=\text{ }\left( 5\text{ }+\text{ }2 \right)7k\text{ }\text{ }2\left( 2k \right)\]

    \[=\text{ }5\left( 7k \right)\text{ }+\text{ }2\text{ }\left( 7k\text{ }\text{ }2k \right)\]

    \[=\text{ }5\left( 7k \right)\text{ }+\text{ }2\text{ }\left( 5x \right)\]

Which is distinguishable by 5.

    \[\Rightarrow P\left( k+1 \right)\]

is valid when P(k) is valid.

In this manner, by Mathematical Induction,

    \[P(n)={{7}^{n}}~\text{ }{{2}^{n}}~\]

is separable by 5 is valid for every regular number n.

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