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Home » Prove n^3 – 7n + 3 is divisible by 3, for all natural numbers n.
Prove n^3 – 7n + 3 is divisible by 3, for all natural numbers n.
CBSE | Class 11 | Maths | NCERT Exemplar | Principle of Mathematical Induction
Prove n^3 – 7n + 3 is divisible by 3, for all natural numbers n.

As per the inquiry,

    \[P\left( n \right)\text{ }=\text{ }n3\text{ }\text{ }7n\text{ }+\text{ }3\]

is distinguishable by 3.

Along these lines, subbing various qualities for n, we get,

    \[P\left( 0 \right)\text{ }=\text{ }03\text{ }\text{ }7\times 0\text{ }+\text{ }3\text{ }=\text{ }3\]

which is distinguishable by 3.

    \[P\left( 1 \right)\text{ }=\text{ }13\text{ }\text{ }7\times 1\text{ }+\text{ }3\text{ }=\text{ }-3\]

which is separable by 3.

    \[P\left( 2 \right)\text{ }=\text{ }23\text{ }\text{ }7\times 2\text{ }+\text{ }3\text{ }=\text{ }-3\]

which is detachable by 3.

    \[~P\left( 3 \right)\text{ }=\text{ }33\text{ }\text{ }7\times 3\text{ }+\text{ }3\text{ }=\text{ }9\]

which is separable by 3.

Let

    \[P\left( k \right)\text{ }=\text{ }k3\text{ }\text{ }7k\text{ }+\text{ }3\]

be distinct by 3

In this way, we get,

    \[\Rightarrow k37k\text{ }+\text{ }3\text{ }=\text{ }3x.\]

Presently, we likewise get that,

    \[\Rightarrow P\left( k+1 \right)\text{ }=\text{ }\left( k+1 \right)37\left( k+1 \right)\text{ }+\text{ }3\]

    \[=\text{ }k3\text{ }+\text{ }3k2\text{ }+\text{ }3k\text{ }+\text{ }1\text{ }\text{ }7k\text{ }\text{ }7\text{ }+\text{ }3\]

    \[=\text{ }3x\text{ }+\text{ }3\left( k2\text{ }+\text{ }k\text{ }\text{ }2 \right)\]

is detachable by 3.

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

is valid when P(k) is valid.

Subsequently, by Mathematical Induction,

    \[P\left( n \right)\text{ }=\text{ }n3\text{ }\text{ }7n\text{ }+\text{ }3\]

is detachable by 3, for all regular numbers n.

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