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Home » 1 + 2 + 3 + … + n = n (n +1)/2 i.e., the sum of the first n natural numbers is n (n + 1)/2.
1 + 2 + 3 + … + n = n (n +1)/2 i.e., the sum of the first n natural numbers is n (n + 1)/2.
CBSE | Class 11 | Exercise 12.2 | Mathematical Induction | Maths | RD Sharma
1 + 2 + 3 + … + n = n (n +1)/2 i.e., the sum of the first n natural numbers is n (n + 1)/2.

Let,

    \[P\text{ }\left( n \right)\text{ }=\text{ }1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\ldots ..\text{ }+\text{ }n\text{ }=\text{ }n\text{ }\left( n\text{ }+1 \right)/2\]

For,

    \[n\text{ }=\text{ }1\]

    \[LHS\text{ }of\text{ }P\text{ }\left( n \right)\text{ }=\text{ }1\]

    \[RHS\text{ }of\text{ }P\text{ }\left( n \right)\text{ }=~1\text{ }\left( 1+1 \right)/2\text{ }=\text{ }1\]

So,

    \[LHS\text{ }=\text{ }RHS\]

As, P (n) is true for

    \[n\text{ }=\text{ }1\]

Let P (n) be the true for

    \[n\text{ }=\text{ }k\]

, so

    \[1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\ldots .\text{ }+\text{ }k\text{ }=\text{ }k\text{ }\left( k+1 \right)/2\text{ }\ldots \text{ }\left( i \right)\]

Now,

    \[\left( 1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\ldots \text{ }+\text{ }k \right)\text{ }+\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }=\text{ }k\text{ }\left( k+1 \right)/2\text{ }+\text{ }\left( k+1 \right)\]

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

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

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

P (n) is true for

    \[n\text{ }=\text{ }k\text{ }+\text{ }1\]

P (n) is true for all

    \[n\in N\]

So, by the principle of Mathematical Induction

Hence,

    \[P\text{ }\left( n \right)\text{ }=\text{ }1\text{ }+\text{ }2\text{ }+\text{ }3\text{ }+\text{ }\ldots ..\text{ }+\text{ }n\text{ }=\text{ }n\text{ }\left( n\text{ }+1 \right)/2\]

is true for all n ∈ N.

i

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