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Prove n(n^2 + 5) is divisible by 6, for each natural number n.
CBSE | Class 11 | Maths | NCERT Exemplar | Principle of Mathematical Induction
Prove n(n^2 + 5) is divisible by 6, for each natural number n.

As per the inquiry,

    \[P\left( n \right)\text{ }=\text{ }n\left( n^2\text{ }+\text{ }5 \right)\]

is distinguishable by 6.

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

    \[P\left( 0 \right)\text{ }=\text{ }0\left( 0^2\text{ }+\text{ }5 \right)\text{ }=\text{ }0\]

Which is separable by 6.

    \[P\left( 1 \right)\text{ }=\text{ }1\left( 1^2\text{ }+\text{ }5 \right)\text{ }=\text{ }6\]

Which is separable by 6.

    \[P\left( 2 \right)\text{ }=\text{ }2\left( 2^2\text{ }+\text{ }5 \right)\text{ }=\text{ }18\]

Which is separable by 6.

    \[P\left( 3 \right)\text{ }=\text{ }3\left( 3^2\text{ }+\text{ }5 \right)\text{ }=\text{ }42\]

Which is separable by 6.

Let

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

be separable by 6.

Thus, we get,

    \[\Rightarrow k\left( k^2\text{ }+\text{ }5 \right)\text{ }=\text{ }6x.\]

Presently, we additionally get that,

    \[\Rightarrow P\left( k+1 \right)\text{ }=\text{ }\left( k+1 \right)\left( \left( k+1 \right)^2\text{ }+\text{ }5 \right)\text{ }=\text{ }\left( k+1 \right)\left( k^2+2k+6 \right)\]

    \[=\text{ }k^3\text{ }+\text{ }3k^2\text{ }+\text{ }8k\text{ }+\text{ }6\]

    \[=\text{ }6x+3k^2+3k+6\]

    \[=\text{ }6x+3k\left( k+1 \right)+6[n\left( n+1 \right)\]

is in every case even and separable by 2]

    \[=\text{ }6x\text{ }+\text{ }3\times 2y\text{ }+\text{ }6\]

Which is detachable by 6.

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

is valid when P(k) is valid.

In this manner, by Mathematical Induction,

    \[P\left( n \right)\text{ }=\text{ }n\left( n^2\text{ }+\text{ }5 \right)\]

is separable by 6, for every normal number n.

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