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Prove 2 + 4 + 6 + …+ 2n = n^2 + n for all natural numbers n.
CBSE | Maths | NCERT Exemplar | Principle of Mathematical Induction
Prove 2 + 4 + 6 + …+ 2n = n^2 + n for all natural numbers n.

 

 

As indicated by the inquiry,

    \[P\left( n \right)\text{ }is\text{ }2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\text{ }\ldots \text{ }+\text{ }2n\text{ }=\text{ }n^2\text{ }+\text{ }n.\]

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

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

Which is valid.

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

Which is valid.

    \[P\left( 2 \right)\text{ }=\text{ }2\text{ }+\text{ }4\text{ }=\text{ }2^2\text{ }+\text{ }2\]

Which is valid.

    \[P\left( 3 \right)\text{ }=\text{ }2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }=\text{ }3^2\text{ }+\text{ }2\]

Which is valid.

Let

    \[P\left( k \right)\text{ }=\text{ }2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\text{ }\ldots \text{ }+\text{ }2k\text{ }=\text{ }k^2\text{ }+\text{ }k\]

be valid;

Along these lines, we get,

    \[\Rightarrow P\left( k+1 \right)\text{ }is\text{ }2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\ldots +\text{ }2k\text{ }+\text{ }2\left( k+1 \right)\text{ }=\text{ }k^2\text{ }+\text{ }k\text{ }+\text{ }2k\text{ }+2\]

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

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

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

is valid when P(k) is valid.

Subsequently, by Mathematical Induction,

    \[~2\text{ }+\text{ }4\text{ }+\text{ }6\text{ }+\text{ }\ldots \text{ }+\text{ }2n\text{ }=\text{ }n^2\text{ }+\text{ }n\]

is valid for all normal numbers n.

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